x^2+12x-8=40

Solution for x^2+12x-8=40 equation:

x^2+12x-8=40

We move all terms to the left:

x^2+12x-8-(40)=0

We add all the numbers together, and all the variables

x^2+12x-48=0

a = 1; b = 12; c = -48;
Δ = b2-4ac
Δ = 122-4·1·(-48)
Δ = 336
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:

$\sqrt{\Delta}=\sqrt{336}=\sqrt{16*21}=\sqrt{16}*\sqrt{21}=4\sqrt{21}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-4\sqrt{21}}{2*1}=\frac{-12-4\sqrt{21}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+4\sqrt{21}}{2*1}=\frac{-12+4\sqrt{21}}{2}
$