2y^2+8y+3=0

Solution for 2y^2+8y+3=0 equation:

2y^2+8y+3=0

a = 2; b = 8; c = +3;
Δ = b2-4ac
Δ = 82-4·2·3
Δ = 40
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{40}=\sqrt{4*10}=\sqrt{4}*\sqrt{10}=2\sqrt{10}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-2\sqrt{10}}{2*2}=\frac{-8-2\sqrt{10}}{4}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+2\sqrt{10}}{2*2}=\frac{-8+2\sqrt{10}}{4}
$