96=2x+(3x)2

Solution for 96=2x+(3x)2 equation:

96=2x+(3x)2

We move all terms to the left:

96-(2x+(3x)2)=0

We add all the numbers together, and all the variables

-(+2x+3x^2)+96=0

We get rid of parentheses

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

a = -3; b = -2; c = +96;
Δ = b2-4ac
Δ = -22-4·(-3)·96
Δ = 1156
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1156}=34$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-34}{2*-3}=\frac{-32}{-6}
=5+1/3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+34}{2*-3}=\frac{36}{-6}
=-6
$