2a^2+12a+8=0

Solution for 2a^2+12a+8=0 equation:

2a^2+12a+8=0

a = 2; b = 12; c = +8;
Δ = b2-4ac
Δ = 122-4·2·8
Δ = 80
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{80}=\sqrt{16*5}=\sqrt{16}*\sqrt{5}=4\sqrt{5}$
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-4\sqrt{5}}{2*2}=\frac{-12-4\sqrt{5}}{4}
$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+4\sqrt{5}}{2*2}=\frac{-12+4\sqrt{5}}{4}
$