-5t^2+70t=35

Solution for -5t^2+70t=35 equation:

-5t^2+70t=35

We move all terms to the left:

-5t^2+70t-(35)=0

a = -5; b = 70; c = -35;
Δ = b2-4ac
Δ = 702-4·(-5)·(-35)
Δ = 4200
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:

$\sqrt{\Delta}=\sqrt{4200}=\sqrt{100*42}=\sqrt{100}*\sqrt{42}=10\sqrt{42}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(70)-10\sqrt{42}}{2*-5}=\frac{-70-10\sqrt{42}}{-10}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(70)+10\sqrt{42}}{2*-5}=\frac{-70+10\sqrt{42}}{-10}
$