3x^2+96=36x

Solution for 3x^2+96=36x equation:

3x^2+96=36x

We move all terms to the left:

3x^2+96-(36x)=0

a = 3; b = -36; c = +96;
Δ = b2-4ac
Δ = -362-4·3·96
Δ = 144
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{144}=12$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-36)-12}{2*3}=\frac{24}{6}
=4
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-36)+12}{2*3}=\frac{48}{6}
=8
$