-16t^2+25t=5

Solution for -16t^2+25t=5 equation:

-16t^2+25t=5

We move all terms to the left:

-16t^2+25t-(5)=0

a = -16; b = 25; c = -5;
Δ = b2-4ac
Δ = 252-4·(-16)·(-5)
Δ = 305
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(25)-\sqrt{305}}{2*-16}=\frac{-25-\sqrt{305}}{-32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(25)+\sqrt{305}}{2*-16}=\frac{-25+\sqrt{305}}{-32}
$