t(8t+5)=0

Solution for t(8t+5)=0 equation:

t(8t+5)=0

We multiply parentheses

8t^2+5t=0

a = 8; b = 5; c = 0;
Δ = b2-4ac
Δ = 52-4·8·0
Δ = 25
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{25}=5$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(5)-5}{2*8}=\frac{-10}{16}
=-5/8
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+5}{2*8}=\frac{0}{16}
=0
$