3t^2+8t-8=0

Solution for 3t^2+8t-8=0 equation:

3t^2+8t-8=0

a = 3; b = 8; c = -8;
Δ = b2-4ac
Δ = 82-4·3·(-8)
Δ = 160
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:

$\sqrt{\Delta}=\sqrt{160}=\sqrt{16*10}=\sqrt{16}*\sqrt{10}=4\sqrt{10}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-4\sqrt{10}}{2*3}=\frac{-8-4\sqrt{10}}{6}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+4\sqrt{10}}{2*3}=\frac{-8+4\sqrt{10}}{6}
$