25t-5t^2+30=0

Solution for 25t-5t^2+30=0 equation:

25t-5t^2+30=0

a = -5; b = 25; c = +30;
Δ = b2-4ac
Δ = 252-4·(-5)·30
Δ = 1225
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1225}=35$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(25)-35}{2*-5}=\frac{-60}{-10}
=+6
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(25)+35}{2*-5}=\frac{10}{-10}
=-1
$