# How do you find the roots, real and imaginary, of y= (3x+4)^2-13x-x^2+14  using the quadratic formula?

Sep 10, 2017

THere are two complex conjugate roots:

$x = - \frac{11}{16} \pm \frac{\sqrt{839}}{16} i$

#### Explanation:

We have:

$y = {\left(3 x + 4\right)}^{2} - 13 x - {x}^{2} + 14$

We seek the roots of $y = 0$, ie of:

${\left(3 x + 4\right)}^{2} - 13 x - {x}^{2} + 14 = 0$

Expanding we get:

$9 {x}^{2} + 24 x + 16 - 13 x - {x}^{2} + 14 = 0$
$\therefore 8 {x}^{2} + 11 x + 30 = 0$

$a = 8 , b = 11 , c = 30$
$x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$
$\setminus \setminus = \frac{- 11 \pm \sqrt{{11}^{2} - 4.8 .30}}{2.8}$
$\setminus \setminus = \frac{- 11 \pm \sqrt{121 - 960}}{16}$
$\setminus \setminus = \frac{- 11 \pm \sqrt{- 839}}{16}$
$\setminus \setminus = - \frac{11}{16} \pm \frac{\sqrt{839}}{16} i$