# How do you find the roots, real and imaginary, of y=4x^2 -7x -(x-2)^2  using the quadratic formula?

Apr 13, 2016

$x = \frac{3 + \sqrt{57}}{6}$
$x = \frac{3 - \sqrt{57}}{6}$

#### Explanation:

Given -

$y = 4 {x}^{2} - 7 x - {\left(x - 2\right)}^{2}$

Simplify

$y = 4 {x}^{2} - 7 x - \left({x}^{2} - 4 x + 4\right)$
$y = {4}^{2} - 7 x - {x}^{2} + 4 x - 4$
$y = 3 {x}^{2} - 3 x - 4$

It is in a quadratic form -

Then If $\sqrt{{b}^{2} - \left(4 a c\right)}$ is positive, its roots are real otherwise imaginary.

sqrt((-3)^2-(4 xx 3 xx (-4))
sqrt(9-(-48)
$\sqrt{9 + 48}$
$\sqrt{57}$

Since $\sqrt{57}$ is positive, the roots are real

$x = \frac{- \left(- 3\right) \pm \sqrt{57}}{2 \times 3}$

$x = \frac{3 + \sqrt{57}}{6}$
$x = \frac{3 - \sqrt{57}}{6}$