x^2=3x+36

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

x^2=3x+36

We move all terms to the left:

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

We get rid of parentheses

x^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}
$