t^2+70t=435

Solution for t^2+70t=435 equation:

t^2+70t=435

We move all terms to the left:

t^2+70t-(435)=0

a = 1; b = 70; c = -435;
Δ = b2-4ac
Δ = 702-4·1·(-435)
Δ = 6640
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:

$\sqrt{\Delta}=\sqrt{6640}=\sqrt{16*415}=\sqrt{16}*\sqrt{415}=4\sqrt{415}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(70)-4\sqrt{415}}{2*1}=\frac{-70-4\sqrt{415}}{2}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(70)+4\sqrt{415}}{2*1}=\frac{-70+4\sqrt{415}}{2}
$