(2x^2)+16x+3=0

Solution for (2x^2)+16x+3=0 equation:

(2x^2)+16x+3=0

a = 2; b = 16; c = +3;
Δ = b2-4ac
Δ = 162-4·2·3
Δ = 232
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:

$\sqrt{\Delta}=\sqrt{232}=\sqrt{4*58}=\sqrt{4}*\sqrt{58}=2\sqrt{58}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-2\sqrt{58}}{2*2}=\frac{-16-2\sqrt{58}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+2\sqrt{58}}{2*2}=\frac{-16+2\sqrt{58}}{4}
$