5t(t+10)=0

Solution for 5t(t+10)=0 equation:

5t(t+10)=0

We multiply parentheses

5t^2+50t=0

a = 5; b = 50; c = 0;
Δ = b2-4ac
Δ = 502-4·5·0
Δ = 2500
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{2500}=50$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(50)-50}{2*5}=\frac{-100}{10}
=-10
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(50)+50}{2*5}=\frac{0}{10}
=0
$