4y^2+16y+3=0

Solution for 4y^2+16y+3=0 equation:

4y^2+16y+3=0

a = 4; b = 16; c = +3;
Δ = b2-4ac
Δ = 162-4·4·3
Δ = 208
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:

$\sqrt{\Delta}=\sqrt{208}=\sqrt{16*13}=\sqrt{16}*\sqrt{13}=4\sqrt{13}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-4\sqrt{13}}{2*4}=\frac{-16-4\sqrt{13}}{8}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+4\sqrt{13}}{2*4}=\frac{-16+4\sqrt{13}}{8}
$