1.25t^2-15t=0

Solution for 1.25t^2-15t=0 equation:

1.25t^2-15t=0

a = 1.25; b = -15; c = 0;
Δ = b2-4ac
Δ = -152-4·1.25·0
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*1.25}=\frac{0}{2.5}
=0
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-15)+15}{2*1.25}=\frac{30}{2.5}
=12
$