5y^2+6y-4=0

Solution for 5y^2+6y-4=0 equation:

5y^2+6y-4=0

a = 5; b = 6; c = -4;
Δ = b2-4ac
Δ = 62-4·5·(-4)
Δ = 116
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:

$\sqrt{\Delta}=\sqrt{116}=\sqrt{4*29}=\sqrt{4}*\sqrt{29}=2\sqrt{29}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-2\sqrt{29}}{2*5}=\frac{-6-2\sqrt{29}}{10}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+2\sqrt{29}}{2*5}=\frac{-6+2\sqrt{29}}{10}
$