2a^2+16a+12=0

Solution for 2a^2+16a+12=0 equation:

2a^2+16a+12=0

a = 2; b = 16; c = +12;
Δ = b2-4ac
Δ = 162-4·2·12
Δ = 160
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:

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