2w^2+3w=160

Solution for 2w^2+3w=160 equation:

2w^2+3w=160

We move all terms to the left:

2w^2+3w-(160)=0

a = 2; b = 3; c = -160;
Δ = b2-4ac
Δ = 32-4·2·(-160)
Δ = 1289
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-\sqrt{1289}}{2*2}=\frac{-3-\sqrt{1289}}{4}
$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+\sqrt{1289}}{2*2}=\frac{-3+\sqrt{1289}}{4}
$