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

Aug 23, 2017

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

#### Explanation:

Solve:

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

Expand ${\left(x - 2\right)}^{2}$.

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

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

Simplify.

$y = 4 {x}^{2} - 7 x - {x}^{2} + 4 x - 4$

Gather like terms.

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

Combine like terms.

$y = 3 {x}^{2} - 3 x - 4$

To solve for roots, substitute $0$ for $y$. Then solve for $x$ using the quadratic formula.

$0 = 3 {x}^{2} - 3 x - 4$

$x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

Plug in the given values.

$x = \frac{- \left(- 3\right) \pm \sqrt{{\left(- 3\right)}^{2} - 4 \cdot 3 \cdot - 4}}{2 \cdot 3}$

Simplify.

$x = \frac{3 \pm \sqrt{9 + 48}}{6}$

$x = \frac{3 \pm \sqrt{57}}{12}$ $\leftarrow$ $57$ is the product of two prime numbers.

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