0=-5t2+35t+5

Solution for 0=-5t2+35t+5 equation:

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

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We get rid of parentheses

5t^2-35t-5=0

a = 5; b = -35; c = -5;
Δ = b2-4ac
Δ = -352-4·5·(-5)
Δ = 1325
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:

$\sqrt{\Delta}=\sqrt{1325}=\sqrt{25*53}=\sqrt{25}*\sqrt{53}=5\sqrt{53}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-35)-5\sqrt{53}}{2*5}=\frac{35-5\sqrt{53}}{10}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-35)+5\sqrt{53}}{2*5}=\frac{35+5\sqrt{53}}{10}
$