-3=x^2+13(x)+8

Solution for -3=x^2+13(x)+8 equation:

-3=x^2+13(x)+8

We move all terms to the left:

-3-(x^2+13(x)+8)=0

We get rid of parentheses

-x^2-13x-8-3=0

We add all the numbers together, and all the variables

-1x^2-13x-11=0

a = -1; b = -13; c = -11;
Δ = b2-4ac
Δ = -132-4·(-1)·(-11)
Δ = 125
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:

$\sqrt{\Delta}=\sqrt{125}=\sqrt{25*5}=\sqrt{25}*\sqrt{5}=5\sqrt{5}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-13)-5\sqrt{5}}{2*-1}=\frac{13-5\sqrt{5}}{-2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-13)+5\sqrt{5}}{2*-1}=\frac{13+5\sqrt{5}}{-2}
$