# How do you solve the quadratic using the quadratic formula given b^2-4b-14=-2?

Aug 12, 2016

Let the equation be ${x}^{2} - 4 x - 14 = - 2$ so that the variables in our equation don't conflict with the parameters in the quadratic formula.

The quadratic formula is $x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$, for equations of the form $a {x}^{2} + b x + c = 0$.

Sending all our terms to one side:

${x}^{2} - 4 x - 14 = - 2$

${x}^{2} - 4 x - 14 + 2 = 0$

${x}^{2} - 4 x - 12 = 0$

Applying the formula:

$x = \frac{- \left(- 4\right) \pm \sqrt{{\left(- 4\right)}^{2} - \left(4 \times 1 \times - 12\right)}}{2 \times 1}$

$x = \frac{4 \pm \sqrt{16 + 48}}{2}$

$x = \frac{4 \pm \sqrt{64}}{2}$

$x = \frac{4 \pm 8}{2}$

$x = \frac{4 + 8}{2} \mathmr{and} \frac{4 - 8}{2}$

$x = \frac{12}{2} \mathmr{and} - \frac{4}{2}$

$x = 6 \mathmr{and} - 2$

Checking in the original equation , we have that both solutions work!

Therefore, the solution set is $\left\{x = 6 \text{ & } - 2\right\}$.

Hopefully this helps!