# How do you find the roots, real and imaginary, of y= (3/4x+1460)x+300  using the quadratic formula?

Mar 15, 2018

$x = \frac{- 2920 \pm 20 \sqrt{21307}}{3}$

#### Explanation:

We take $\left(\frac{3}{4} x + 1460\right) x + 300$ and expand it:

$\frac{3}{4} {x}^{2} + 1460 x + 300$

We equate this to $0$, and use the quadratic formula to solve it.

The formula states that:

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

Here, $a = \frac{3}{4} , b = 1460 , c = 300$. Inputting:

$x = \frac{- 1460 \pm \sqrt{{1460}^{2} - 4 \left(\frac{3}{4}\right) \left(300\right)}}{2 \cdot \frac{3}{4}}$

$x = \frac{- 1460 \pm \sqrt{2131600 - 900}}{\frac{3}{2}}$

$x = \frac{- 1460 \pm \sqrt{2130700}}{\frac{3}{2}}$

$x = \frac{- 2920 \pm 20 \sqrt{21307}}{3}$