4w^2+6w=160

Solution for 4w^2+6w=160 equation:

4w^2+6w=160

We move all terms to the left:

4w^2+6w-(160)=0

a = 4; b = 6; c = -160;
Δ = b2-4ac
Δ = 62-4·4·(-160)
Δ = 2596
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{2596}=\sqrt{4*649}=\sqrt{4}*\sqrt{649}=2\sqrt{649}$
$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-2\sqrt{649}}{2*4}=\frac{-6-2\sqrt{649}}{8}
$
$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+2\sqrt{649}}{2*4}=\frac{-6+2\sqrt{649}}{8}
$