35=-5t^2+20t+25

Solution for 35=-5t^2+20t+25 equation:

35=-5t^2+20t+25

We move all terms to the left:

35-(-5t^2+20t+25)=0

We get rid of parentheses

5t^2-20t-25+35=0

We add all the numbers together, and all the variables

5t^2-20t+10=0

a = 5; b = -20; c = +10;
Δ = b2-4ac
Δ = -202-4·5·10
Δ = 200
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{200}=\sqrt{100*2}=\sqrt{100}*\sqrt{2}=10\sqrt{2}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-10\sqrt{2}}{2*5}=\frac{20-10\sqrt{2}}{10}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+10\sqrt{2}}{2*5}=\frac{20+10\sqrt{2}}{10}
$