-35=35-5t-5t^2

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

-35=35-5t-5t^2

We move all terms to the left:

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

5t^2+5t-70=0

a = 5; b = 5; c = -70;
Δ = b2-4ac
Δ = 52-4·5·(-70)
Δ = 1425
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:

$\sqrt{\Delta}=\sqrt{1425}=\sqrt{25*57}=\sqrt{25}*\sqrt{57}=5\sqrt{57}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(5)-5\sqrt{57}}{2*5}=\frac{-5-5\sqrt{57}}{10}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+5\sqrt{57}}{2*5}=\frac{-5+5\sqrt{57}}{10}
$