2x2+16x+8=0

Solution for 2x2+16x+8=0 equation:

2x^2+16x+8=0

a = 2; b = 16; c = +8;
Δ = b2-4ac
Δ = 162-4·2·8
Δ = 192
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{192}=\sqrt{64*3}=\sqrt{64}*\sqrt{3}=8\sqrt{3}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-8\sqrt{3}}{2*2}=\frac{-16-8\sqrt{3}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+8\sqrt{3}}{2*2}=\frac{-16+8\sqrt{3}}{4}
$