2t^2+8t-4=0

Solution for 2t^2+8t-4=0 equation:

2t^2+8t-4=0

a = 2; b = 8; c = -4;
Δ = b2-4ac
Δ = 82-4·2·(-4)
Δ = 96
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:

$\sqrt{\Delta}=\sqrt{96}=\sqrt{16*6}=\sqrt{16}*\sqrt{6}=4\sqrt{6}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-4\sqrt{6}}{2*2}=\frac{-8-4\sqrt{6}}{4}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+4\sqrt{6}}{2*2}=\frac{-8+4\sqrt{6}}{4}
$