5t^2+24t=0

Solution for 5t^2+24t=0 equation:

5t^2+24t=0

a = 5; b = 24; c = 0;
Δ = b2-4ac
Δ = 242-4·5·0
Δ = 576
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{576}=24$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(24)-24}{2*5}=\frac{-48}{10}
=-4+4/5
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(24)+24}{2*5}=\frac{0}{10}
=0
$