4y^2+12y-3=0

Solution for 4y^2+12y-3=0 equation:

4y^2+12y-3=0

a = 4; b = 12; c = -3;
Δ = b2-4ac
Δ = 122-4·4·(-3)
Δ = 192
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:

$\sqrt{\Delta}=\sqrt{192}=\sqrt{64*3}=\sqrt{64}*\sqrt{3}=8\sqrt{3}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-8\sqrt{3}}{2*4}=\frac{-12-8\sqrt{3}}{8}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+8\sqrt{3}}{2*4}=\frac{-12+8\sqrt{3}}{8}
$