0=3x2+36x+13

Solution for 0=3x2+36x+13 equation:

0=3x^2+36x+13

We move all terms to the left:

0-(3x^2+36x+13)=0

We add all the numbers together, and all the variables

-(3x^2+36x+13)=0

We get rid of parentheses

-3x^2-36x-13=0

a = -3; b = -36; c = -13;
Δ = b2-4ac
Δ = -362-4·(-3)·(-13)
Δ = 1140
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:

$\sqrt{\Delta}=\sqrt{1140}=\sqrt{4*285}=\sqrt{4}*\sqrt{285}=2\sqrt{285}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-36)-2\sqrt{285}}{2*-3}=\frac{36-2\sqrt{285}}{-6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-36)+2\sqrt{285}}{2*-3}=\frac{36+2\sqrt{285}}{-6}
$