X^2+36=2x^2+3x

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

X^2+36=2X^2+3X

We move all terms to the left:

X^2+36-(2X^2+3X)=0

We get rid of parentheses

X^2-2X^2-3X+36=0

We add all the numbers together, and all the variables

-1X^2-3X+36=0

a = -1; b = -3; c = +36;
Δ = b2-4ac
Δ = -32-4·(-1)·36
Δ = 153
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:

$\sqrt{\Delta}=\sqrt{153}=\sqrt{9*17}=\sqrt{9}*\sqrt{17}=3\sqrt{17}$
$X_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-3\sqrt{17}}{2*-1}=\frac{3-3\sqrt{17}}{-2}
$
$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+3\sqrt{17}}{2*-1}=\frac{3+3\sqrt{17}}{-2}
$