# How do you solve 8x^2+4=-33x?

Apr 5, 2016

The solutions are:
x = color(blue)(-1/8

x= color(blue)( - 4

#### Explanation:

$8 {x}^{2} + 4 = - 33 x$

$8 {x}^{2} + 4 + 33 x = 0$

$8 {x}^{2} + 33 x + 4 = 0$

The equation is of the form color(blue)(ax^2+bx+c=0 where:

$a = 8 , b = 33 , c = 4$

The Discriminant is given by:

$\Delta = {b}^{2} - 4 \cdot a \cdot c$

$= {\left(33\right)}^{2} - \left(4 \cdot 8 \cdot 4\right)$

$= 1089 - 128 = 961$

The solutions are found using the formula
$x = \frac{- b \pm \sqrt{\Delta}}{2 \cdot a}$

$= \frac{- 33 \pm \sqrt{961}}{2 \cdot 8}$

$= \frac{- 33 \pm 31}{16}$

x =(-33+31)/(16) = -2/16 = color(blue)(-1/8

x =(-33-31)/(16) = -64/16 = color(blue)( - 4

Apr 6, 2016

$y = 8 {x}^{2} + 33 x + 4 = 0$
Transformed equation: $y ' = {x}^{2} + 33 x + 32$.
Back to y, divide these real roots by (a = 8) to get the 2 real roots of y --> : $x 1 = - \frac{1}{8}$ and $x 2 = - \frac{32}{8} = - 4.$