2q^2+160q=0

Solution for 2q^2+160q=0 equation:

2q^2+160q=0

a = 2; b = 160; c = 0;
Δ = b2-4ac
Δ = 1602-4·2·0
Δ = 25600
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{25600}=160$
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(160)-160}{2*2}=\frac{-320}{4}
=-80
$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(160)+160}{2*2}=\frac{0}{4}
=0
$