The quadratic formula is a math formula that can be used to solve a quadratic equation that is written in standard form ax^{2} + bx + c = 0

Notice that the general form of a quadratic equation is ax^{2} + bx + c = 0 and it is a second order equation in a single variable. Before using the formula shown above, it is important to check two things:

- First, make sure that the coefficient of the leading term is not equal to zero (a ≠ 0). If a is equal to zero, then the equation becomes a linear equation instead.

- Then, make sure that the quadratic equation is indeed written in standard or general form.

The plus or minus sign (±) in the formula is there to show that the quadratic equation may have two solutions (x_{1} and x_{2}) generally speaking.

- The expression inside the radical symbol or square root sign is called
**radicand**. Therefore, the radicand in the quadratic formula is b^{2}- 4ac.

- The
**discriminant**of a quadratic equation in the form ax^{2}+ bx + c = 0 is the value of the expression b^{2}- 4ac.

- When you are dealing with a function, x
_{1}and x_{2}are usually called**zeros**of the function.

- When you are dealing with a quadratic equation, x
_{1}and x_{2}are usually called**solutions**or**roots**of the quadratic equation.

- When you are graphing a quadratic function, (x
_{1}, 0) and and (x_{2}, 0) are called x-intercepts since these are points where the parabola crosses the x-axis.

The expression b^{2} - 4ac, called discriminant, shows the nature of the solutions.

It is common to use the symbol Δ (delta from the Greek alphabet) when talking about the discriminant.

Δ = b^{2} - 4ac

If Δ or the discriminant is zero, then it makes no difference whether we choose the plus or the minus sign in the formula.

x_{1} = x_{2} = -b/2a

In this case, we say that there is one repeated real solution.

If Δ or the discriminant is positive, then there will be two real solutions.

If Δ or the discriminant is negative, then we will end up taking the square root of a negative number. In this case, there will be two imaginary-number solutions called complex numbers.

**Discriminant**

For ax^{2} + bx + c = 0:

Δ = b^{2} - 4ac = 0 ⟶ One real-number solution

Δ = b^{2} - 4ac > 0 ⟶ Two different real-number solutions

Δ = b^{2} - 4ac < 0 ⟶ No real solution, but two different imaginary-number solutions.

Check the lesson about discriminant of the quadratic equation to see what the graph of a quadratic equation looks like when the discriminant is either zero, positive, or negative.

Solve x^{2} - 5x + 4 = 0 using the quadratic formula

a = 1, b = -5, and c = 4

$$ x = \frac{-(-5) ± \sqrt{(-5)^2 - 4(1)(4)}}{2(1)} $$ $$ x = \frac{5 ± \sqrt{25 - 16}}{2} $$ $$ x = \frac{5 ± \sqrt{9}}{2} $$ $$ x = \frac{5 ± 3}{2} $$ $$ x_1 = \frac{5 + 3}{2}\ and\ x_2 = \frac{5 - 3}{2} $$ $$ x_1 = \frac{8}{2}\ and\ x_2 = \frac{2}{2} $$ $$ x_1 = 4\ and\ x_2 = 1 $$

The roots of the equation x^{2} - 5x + 4 = 0 are **x _{1} = 4** and

Please check the lesson about solve using the quadratic formula to see more examples.

**Example #1**

Suppose a soccer player shoots a penalty kick with an initial velocity of 28 ft/s. When will the ball reach a height of 30 feet?

**Solution**

The function h = -16t^{2} + vt + s models the height h in feet of the ball at time t in seconds.

The velocity is v and s is the initial height of the ball.

Since the soccer ball must be on the ground before the soccer player shoots the ball, s is equal to 0.

v = 28 ft/s

h is the height of the ball

30 = -16t^{2} + 28t

Since the standard form of a quadratic equation is ax^{2} + bx + c = 0, you need to put 30 = -16t^{2} + 28t in standard form.

Subtract 30 from both sides of the equation

30 - 30 = -16t^{2} + 28t - 30

0 = -16t^{2} + 28t - 30

-16t^{2} + 28t - 30 = 0

Find the values of a,b, and c and then evaluate the discriminant.

a = -16, b = 28 and c = -30

Δ = b^{2} - 4ac = 28^{2} - 4(-16)(-30)

Δ = b^{2} - 4ac = 784 + 64(-30)

Δ = b^{2} - 4ac = 784 + -1920

Δ = b^{2} - 4ac = -1136

Since the discriminant is negative, the equation 30 = -16t^{2} + 28t has no real solutions.

Therefore, the ball will not reach a height of 30 feet.

**Example #2**

Find the dimensions of a square that has the same area as a circle whose radius is 10 inches.

**Solution**

Let x be the length of one side of the square. Then, the area of the square is x^{2}

The area of the circle is pir^{2} = 3.14(10)^{2} = 3.14(100) = 314

x^{2} = 314

x^{2} - 314 = 0

x^{2} - 0x - 314 = 0

a = 1, b = 0, and c = -314

Δ = b^{2} - 4ac = 0^{2} - 4(1)(-314)

Δ = b^{2} - 4ac = 1256

√Δ = √(1256) = 35.44

x_{1} = (-b + 35.44) / 2(1)

x_{1} = (-0 + 35.44) / 2

**x _{1} = 35.44 / 2 = 17.72**

x_{2} = (-b - 35.44) / 2(1)

x_{2} = (-0 - 35.44) / 2

**x _{2} = -35.44 / 2 = -17.72**

The length of one side of the square is 17.72 inches