35=-5t^2+30t

Solution for 35=-5t^2+30t equation:

35=-5t^2+30t

We move all terms to the left:

35-(-5t^2+30t)=0

We get rid of parentheses

5t^2-30t+35=0

a = 5; b = -30; c = +35;
Δ = b2-4ac
Δ = -302-4·5·35
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{-(-30)-10\sqrt{2}}{2*5}=\frac{30-10\sqrt{2}}{10}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-30)+10\sqrt{2}}{2*5}=\frac{30+10\sqrt{2}}{10}
$