-16t^2+50t=3

Solution for -16t^2+50t=3 equation:

-16t^2+50t=3

We move all terms to the left:

-16t^2+50t-(3)=0

a = -16; b = 50; c = -3;
Δ = b2-4ac
Δ = 502-4·(-16)·(-3)
Δ = 2308
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:

$\sqrt{\Delta}=\sqrt{2308}=\sqrt{4*577}=\sqrt{4}*\sqrt{577}=2\sqrt{577}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(50)-2\sqrt{577}}{2*-16}=\frac{-50-2\sqrt{577}}{-32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(50)+2\sqrt{577}}{2*-16}=\frac{-50+2\sqrt{577}}{-32}
$