35=-7t^2+28t+35

Solution for 35=-7t^2+28t+35 equation:

35=-7t^2+28t+35

We move all terms to the left:

35-(-7t^2+28t+35)=0

We get rid of parentheses

7t^2-28t-35+35=0

We add all the numbers together, and all the variables

7t^2-28t=0

a = 7; b = -28; c = 0;
Δ = b2-4ac
Δ = -282-4·7·0
Δ = 784
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{784}=28$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-28)-28}{2*7}=\frac{0}{14}
=0
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-28)+28}{2*7}=\frac{56}{14}
=4
$