x^2+2x-5=22

Solution for x^2+2x-5=22 equation:

x^2+2x-5=22

We move all terms to the left:

x^2+2x-5-(22)=0

We add all the numbers together, and all the variables

x^2+2x-27=0

a = 1; b = 2; c = -27;
Δ = b2-4ac
Δ = 22-4·1·(-27)
Δ = 112
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:

$\sqrt{\Delta}=\sqrt{112}=\sqrt{16*7}=\sqrt{16}*\sqrt{7}=4\sqrt{7}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-4\sqrt{7}}{2*1}=\frac{-2-4\sqrt{7}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+4\sqrt{7}}{2*1}=\frac{-2+4\sqrt{7}}{2}
$