3t2-32t-64=0

Solution for 3t2-32t-64=0 equation:

3t^2-32t-64=0

a = 3; b = -32; c = -64;
Δ = b2-4ac
Δ = -322-4·3·(-64)
Δ = 1792
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:

$\sqrt{\Delta}=\sqrt{1792}=\sqrt{256*7}=\sqrt{256}*\sqrt{7}=16\sqrt{7}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-32)-16\sqrt{7}}{2*3}=\frac{32-16\sqrt{7}}{6}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-32)+16\sqrt{7}}{2*3}=\frac{32+16\sqrt{7}}{6}
$