9x2+20x=3x2-34x+108

Solution for 9x2+20x=3x2-34x+108 equation:

9x^2+20x=3x^2-34x+108

We move all terms to the left:

9x^2+20x-(3x^2-34x+108)=0

We get rid of parentheses

9x^2-3x^2+20x+34x-108=0

We add all the numbers together, and all the variables

6x^2+54x-108=0

a = 6; b = 54; c = -108;
Δ = b2-4ac
Δ = 542-4·6·(-108)
Δ = 5508
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:

$\sqrt{\Delta}=\sqrt{5508}=\sqrt{324*17}=\sqrt{324}*\sqrt{17}=18\sqrt{17}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(54)-18\sqrt{17}}{2*6}=\frac{-54-18\sqrt{17}}{12}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(54)+18\sqrt{17}}{2*6}=\frac{-54+18\sqrt{17}}{12}
$