t=-5t^2+30t-25

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

t=-5t^2+30t-25

We move all terms to the left:

t-(-5t^2+30t-25)=0

We get rid of parentheses

5t^2-30t+t+25=0

We add all the numbers together, and all the variables

5t^2-29t+25=0

a = 5; b = -29; c = +25;
Δ = b2-4ac
Δ = -292-4·5·25
Δ = 341
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-29)-\sqrt{341}}{2*5}=\frac{29-\sqrt{341}}{10}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-29)+\sqrt{341}}{2*5}=\frac{29+\sqrt{341}}{10}
$