-32t^2+70t+11=0

Solution for -32t^2+70t+11=0 equation:

-32t^2+70t+11=0

a = -32; b = 70; c = +11;
Δ = b2-4ac
Δ = 702-4·(-32)·11
Δ = 6308
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:

$\sqrt{\Delta}=\sqrt{6308}=\sqrt{4*1577}=\sqrt{4}*\sqrt{1577}=2\sqrt{1577}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(70)-2\sqrt{1577}}{2*-32}=\frac{-70-2\sqrt{1577}}{-64}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(70)+2\sqrt{1577}}{2*-32}=\frac{-70+2\sqrt{1577}}{-64}
$