4y^2+6y-3=0

Solution for 4y^2+6y-3=0 equation:

4y^2+6y-3=0

a = 4; b = 6; c = -3;
Δ = b2-4ac
Δ = 62-4·4·(-3)
Δ = 84
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:

$\sqrt{\Delta}=\sqrt{84}=\sqrt{4*21}=\sqrt{4}*\sqrt{21}=2\sqrt{21}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-2\sqrt{21}}{2*4}=\frac{-6-2\sqrt{21}}{8}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+2\sqrt{21}}{2*4}=\frac{-6+2\sqrt{21}}{8}
$