8t^2+15t=0

Solution for 8t^2+15t=0 equation:

8t^2+15t=0

a = 8; b = 15; c = 0;
Δ = b2-4ac
Δ = 152-4·8·0
Δ = 225
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{225}=15$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(15)-15}{2*8}=\frac{-30}{16}
=-1+7/8
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(15)+15}{2*8}=\frac{0}{16}
=0
$