12x+x^2=160

Solution for 12x+x^2=160 equation:

12x+x^2=160

We move all terms to the left:

12x+x^2-(160)=0

a = 1; b = 12; c = -160;
Δ = b2-4ac
Δ = 122-4·1·(-160)
Δ = 784
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{784}=28$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-28}{2*1}=\frac{-40}{2}
=-20
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+28}{2*1}=\frac{16}{2}
=8
$