32t^2-20t-3=0

Solution for 32t^2-20t-3=0 equation:

32t^2-20t-3=0

a = 32; b = -20; c = -3;
Δ = b2-4ac
Δ = -202-4·32·(-3)
Δ = 784
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{784}=28$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-28}{2*32}=\frac{-8}{64}
=-1/8
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+28}{2*32}=\frac{48}{64}
=3/4
$