160+4x=2x^2

Solution for 160+4x=2x^2 equation:

160+4x=2x^2

We move all terms to the left:

160+4x-(2x^2)=0

determiningTheFunctionDomain
-2x^2+4x+160=0

a = -2; b = 4; c = +160;
Δ = b2-4ac
Δ = 42-4·(-2)·160
Δ = 1296
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1296}=36$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-36}{2*-2}=\frac{-40}{-4}
=+10
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+36}{2*-2}=\frac{32}{-4}
=-8
$