-16t^2+10t=-3

Solution for -16t^2+10t=-3 equation:

-16t^2+10t=-3

We move all terms to the left:

-16t^2+10t-(-3)=0

We add all the numbers together, and all the variables

-16t^2+10t+3=0

a = -16; b = 10; c = +3;
Δ = b2-4ac
Δ = 102-4·(-16)·3
Δ = 292
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:

$\sqrt{\Delta}=\sqrt{292}=\sqrt{4*73}=\sqrt{4}*\sqrt{73}=2\sqrt{73}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-2\sqrt{73}}{2*-16}=\frac{-10-2\sqrt{73}}{-32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+2\sqrt{73}}{2*-16}=\frac{-10+2\sqrt{73}}{-32}
$