The Quadratic Formula

The solutions to the quadratic equation of the form

$$a{x^2} + bx + c = 0$$

where \(a \ne 0\) are given by the formula

$$x = {{ – b \pm \sqrt {{b^2} – 4ac} } \over {2a}}$$

Example 1: Solve the quadratic equation \(3{x^2} – 2x – 5 = 0\).

From the equation \(3{x^2} – 2x – 5 = 0\), we extract the following values:

$$\eqalign{
a &= 3 \cr
b &= – 2 \cr
c &= – 5 \cr} $$

We substitute these into the formula and simplify.

$$\eqalign{
x &= {{ – b \pm \sqrt {{b^2} – 4ac} } \over {2a}} \cr
&= {{ – \left( { – 2} \right) \pm \sqrt {{{\left( { – 2} \right)}^2} – 4\left( 3 \right)\left( { – 5} \right)} } \over {2\left( 3 \right)}} \cr
&= {{2 \pm \sqrt {4 + 60} } \over 6} \cr
&= {{2 \pm \sqrt {64} } \over 6} \cr
x &= {{2 \pm 8} \over 6} \cr} $$

Consider the plus or minus case.

$${x_1} = {{2 + 8} \over 6} = {{10} \over 6} = {5 \over 3}$$

$${x_2} = {{2 – 8} \over 6} = {{ – 6} \over 6} = \,- 1$$

Example 2: Solve \(5{x^2} = 2\, – 4x\).

The quadratic equation is not in the form that we want. We need one side of the equation to equal zero. We can accomplish that by adding both sides by \(4x\) then subtracting both sides by \(2\).

$$5{x^2} + 4x – 2 = 0$$

That means the values are

$$\eqalign{
a &= 5 \cr
b &= 4 \cr
c &= – 2 \cr} $$

Let’s plug in the values of a, b, and c then simplify.

$$\eqalign{
x &= {{ – b \pm \sqrt {{b^2} – 4ac} } \over {2a}} \cr
&= {{ – 4 \pm \sqrt {{4^2} – 4\left( 5 \right)\left( { – 2} \right)} } \over {2\left( 5 \right)}} \cr
&= {{ – 4 \pm \sqrt {16 + 40} } \over {10}} \cr
&= {{ – 4 \pm \sqrt {56} } \over {10}} \cr
&= {{ – 4 \pm \sqrt 4 \sqrt {14} } \over {10}} \cr
&= {{ – 4 \pm 2\sqrt {14} } \over {10}} \cr
x &= {{ – 2 \pm \sqrt {14} } \over 5} \cr} $$

Notice that we factor out 2 in the numerator which we use to cancel the 2 in the denominator.

The solutions are

$$\eqalign{
& {x_1} = {{ – 2 + \sqrt {14} } \over 5} \cr
& {x_2} = {{ – 2 – \sqrt {14} } \over 5} \cr} $$