4x+x^2=96

Solution for 4x+x^2=96 equation:

4x+x^2=96

We move all terms to the left:

4x+x^2-(96)=0

a = 1; b = 4; c = -96;
Δ = b2-4ac
Δ = 42-4·1·(-96)
Δ = 400
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{400}=20$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-20}{2*1}=\frac{-24}{2}
=-12
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+20}{2*1}=\frac{16}{2}
=8
$