2c^2+4c-8=0

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

2c^2+4c-8=0

a = 2; b = 4; c = -8;
Δ = b2-4ac
Δ = 42-4·2·(-8)
Δ = 80
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{80}=\sqrt{16*5}=\sqrt{16}*\sqrt{5}=4\sqrt{5}$
$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-4\sqrt{5}}{2*2}=\frac{-4-4\sqrt{5}}{4}
$
$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+4\sqrt{5}}{2*2}=\frac{-4+4\sqrt{5}}{4}
$