8x+96=2x^2+x^2

Solution for 8x+96=2x^2+x^2 equation:

8x+96=2x^2+x^2

We move all terms to the left:

8x+96-(2x^2+x^2)=0

We get rid of parentheses

-2x^2-x^2+8x+96=0

We add all the numbers together, and all the variables

-3x^2+8x+96=0

a = -3; b = 8; c = +96;
Δ = b2-4ac
Δ = 82-4·(-3)·96
Δ = 1216
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{1216}=\sqrt{64*19}=\sqrt{64}*\sqrt{19}=8\sqrt{19}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-8\sqrt{19}}{2*-3}=\frac{-8-8\sqrt{19}}{-6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+8\sqrt{19}}{2*-3}=\frac{-8+8\sqrt{19}}{-6}
$