# How do you find the roots, real and imaginary, of y=2x^2 + 13x +6  using the quadratic formula?

Mar 16, 2016

real roots: $x = - \frac{1}{2} , - 6$
imaginary roots: none

#### Explanation:

$1$. Since the given equation is already in standard form, identify the $\textcolor{b l u e}{a} , \textcolor{\mathrm{da} r k \mathmr{and} a n \ge}{b} ,$ and $\textcolor{v i o \le t}{c}$ values. Then plug the values into the quadratic formula to solve for the roots.

$y = \textcolor{b l u e}{2} {x}^{2}$ $\textcolor{\mathrm{da} r k \mathmr{and} a n \ge}{+ 13} x$ $\textcolor{v i o \le t}{+ 6}$

$\textcolor{b l u e}{a = 2} \textcolor{w h i t e}{X X X X X} \textcolor{\mathrm{da} r k \mathmr{and} a n \ge}{b = 13} \textcolor{w h i t e}{X X X X X} \textcolor{v i o \le t}{c = 6}$

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

$x = \frac{- \left(\textcolor{\mathrm{da} r k \mathmr{and} a n \ge}{13}\right) \pm \sqrt{{\left(\textcolor{\mathrm{da} r k \mathmr{and} a n \ge}{13}\right)}^{2} - 4 \left(\textcolor{b l u e}{2}\right) \left(\textcolor{v i o \le t}{6}\right)}}{2 \left(\textcolor{b l u e}{2}\right)}$

$x = \frac{- 13 \pm \sqrt{169 - 48}}{4}$

$x = \frac{- 13 \pm \sqrt{121}}{4}$

$x = \frac{- 13 \pm 11}{4}$

$x = \frac{- 13 + 11}{4} \textcolor{w h i t e}{X} , \textcolor{w h i t e}{X} \frac{- 13 - 11}{4}$

$x = - \frac{2}{4} \textcolor{w h i t e}{X} , \textcolor{w h i t e}{X} - \frac{24}{4}$

$\textcolor{g r e e n}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} x = - \frac{1}{2} , - 6 \textcolor{w h i t e}{\frac{a}{a}} |}}}$

$\therefore$, the real roots are $x = - \frac{1}{2}$ or $x = - 6$.