# How do you find the roots, real and imaginary, of y=2x^2 -452x-68  using the quadratic formula?

Dec 20, 2017

Substitute in the values of $a$, $b$ and $c$ into the quadratic formula to get $x = 113 \pm \sqrt{12803}$.

#### Explanation:

We have $y = 2 {x}^{2} - 452 x - 68$, where:

$a = 2$,

$b = - 452$, and

$c = - 68$.

Plug them in to the quadratic formula:

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

$x = \frac{- \left(- 452\right) \pm \sqrt{{\left(- 452\right)}^{2} - 4 \left(2\right) \left(- 68\right)}}{2 \left(2\right)}$

We'll evaluate this starting from what's inside the square root. However, I think we should do some prime factorization.

$452 = {2}^{2} \cdot 113$

${\left(- 452\right)}^{2} = {2}^{4} \cdot {113}^{2}$

Substitute this back in:

$x = \frac{- \left(- 452\right) \pm \sqrt{{2}^{4} \cdot {113}^{2} - 4 \left(2\right) \left(- 68\right)}}{2 \left(2\right)}$

Then do the same for $- 4 \left(2\right) \left(- 68\right)$:

$- 4 \cdot 2 \cdot - 68 = 4 \cdot 2 \cdot 68$

$4 \cdot 2 \cdot 68 = \left({2}^{2}\right) \cdot \left({2}^{1}\right) \cdot \left({2}^{2} \cdot 17\right) = {2}^{5} \cdot 17$

So we have:

$x = \frac{- \left(- 452\right) \pm \sqrt{{2}^{4} \cdot {113}^{2} + {2}^{5} \cdot 17}}{2 \left(2\right)}$

Factor out what's common, which is ${2}^{4}$:

$x = \frac{- \left(- 452\right) \pm \sqrt{{2}^{4} \left({113}^{2} + 2 \cdot 17\right)}}{2 \left(2\right)}$

Then evaluate $\sqrt{{2}^{4}}$:

$x = \frac{- \left(- 452\right) \pm {2}^{2} \sqrt{{113}^{2} + 2 \cdot 17}}{2 \left(2\right)}$

$x = \frac{- \left(- 452\right) \pm 4 \sqrt{{113}^{2} + 2 \cdot 17}}{2 \left(2\right)}$

Now, let's look at what's outside the square root. $- \left(- 452\right)$ becomes $452$, and $2 \left(2\right)$ becomes $4$:

$x = \frac{452 \pm 4 \sqrt{{113}^{2} + 2 \cdot 17}}{4}$

On the numerator, we can factor out $4$:

$x = \frac{4 \left(113 \pm \sqrt{{113}^{2} + 2 \cdot 17}\right)}{4}$

And they cancel out:

$x = 113 \pm \sqrt{{113}^{2} + 2 \cdot 17}$

Let's evaluate what's inside the square root:

${113}^{2} + 2 \cdot 17 = 12769 + 2 \cdot 17 = 12769 + 34 = 12803$

$x = 113 \pm \sqrt{12803}$

So we have two real solutions:

${x}_{1} = 113 + \sqrt{12803}$

${x}_{2} = 113 - \sqrt{12803}$

And no imaginary solutions. We could estimate the value of the square root to get:

${x}_{1} \approx 113 + 113.15 = 226.15$

${x}_{2} \approx 113 - 113.15 = - 0.15$