X^2+22x=-96

Solution for X^2+22x=-96 equation:

X^2+22X=-96

We move all terms to the left:

X^2+22X-(-96)=0

We add all the numbers together, and all the variables

X^2+22X+96=0

a = 1; b = 22; c = +96;
Δ = b2-4ac
Δ = 222-4·1·96
Δ = 100
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$X_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{100}=10$
$X_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(22)-10}{2*1}=\frac{-32}{2}
=-16
$
$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(22)+10}{2*1}=\frac{-12}{2}
=-6
$