3x^2+24x-36=100

Solution for 3x^2+24x-36=100 equation:

3x^2+24x-36=100

We move all terms to the left:

3x^2+24x-36-(100)=0

We add all the numbers together, and all the variables

3x^2+24x-136=0

a = 3; b = 24; c = -136;
Δ = b2-4ac
Δ = 242-4·3·(-136)
Δ = 2208
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:

$\sqrt{\Delta}=\sqrt{2208}=\sqrt{16*138}=\sqrt{16}*\sqrt{138}=4\sqrt{138}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(24)-4\sqrt{138}}{2*3}=\frac{-24-4\sqrt{138}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(24)+4\sqrt{138}}{2*3}=\frac{-24+4\sqrt{138}}{6}
$