x^2+4x-13=5

Solution for x^2+4x-13=5 equation:

x^2+4x-13=5

We move all terms to the left:

x^2+4x-13-(5)=0

We add all the numbers together, and all the variables

x^2+4x-18=0

a = 1; b = 4; c = -18;
Δ = b2-4ac
Δ = 42-4·1·(-18)
Δ = 88
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:

$\sqrt{\Delta}=\sqrt{88}=\sqrt{4*22}=\sqrt{4}*\sqrt{22}=2\sqrt{22}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-2\sqrt{22}}{2*1}=\frac{-4-2\sqrt{22}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+2\sqrt{22}}{2*1}=\frac{-4+2\sqrt{22}}{2}
$