x^2+40x=2304

Solution for x^2+40x=2304 equation:

x^2+40x=2304

We move all terms to the left:

x^2+40x-(2304)=0

a = 1; b = 40; c = -2304;
Δ = b2-4ac
Δ = 402-4·1·(-2304)
Δ = 10816
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{10816}=104$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(40)-104}{2*1}=\frac{-144}{2}
=-72
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(40)+104}{2*1}=\frac{64}{2}
=32
$