3.75t^2-25t+15=0

Solution for 3.75t^2-25t+15=0 equation:

3.75t^2-25t+15=0

a = 3.75; b = -25; c = +15;
Δ = b2-4ac
Δ = -252-4·3.75·15
Δ = 400
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{400}=20$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-25)-20}{2*3.75}=\frac{5}{7.5}
=5/7.5
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-25)+20}{2*3.75}=\frac{45}{7.5}
=6
$