x^2+12x=140

Solution for x^2+12x=140 equation:

x^2+12x=140

We move all terms to the left:

x^2+12x-(140)=0

a = 1; b = 12; c = -140;
Δ = b2-4ac
Δ = 122-4·1·(-140)
Δ = 704
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:

$\sqrt{\Delta}=\sqrt{704}=\sqrt{64*11}=\sqrt{64}*\sqrt{11}=8\sqrt{11}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-8\sqrt{11}}{2*1}=\frac{-12-8\sqrt{11}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+8\sqrt{11}}{2*1}=\frac{-12+8\sqrt{11}}{2}
$