35q^2+24q=0

Solution for 35q^2+24q=0 equation:

35q^2+24q=0

a = 35; b = 24; c = 0;
Δ = b2-4ac
Δ = 242-4·35·0
Δ = 576
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{576}=24$
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(24)-24}{2*35}=\frac{-48}{70}
=-24/35
$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(24)+24}{2*35}=\frac{0}{70}
=0
$