x^2+40x=48

Solution for x^2+40x=48 equation:

x^2+40x=48

We move all terms to the left:

x^2+40x-(48)=0

a = 1; b = 40; c = -48;
Δ = b2-4ac
Δ = 402-4·1·(-48)
Δ = 1792
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:

$\sqrt{\Delta}=\sqrt{1792}=\sqrt{256*7}=\sqrt{256}*\sqrt{7}=16\sqrt{7}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(40)-16\sqrt{7}}{2*1}=\frac{-40-16\sqrt{7}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(40)+16\sqrt{7}}{2*1}=\frac{-40+16\sqrt{7}}{2}
$