4y^2+10y-5=0

Solution for 4y^2+10y-5=0 equation:

4y^2+10y-5=0

a = 4; b = 10; c = -5;
Δ = b2-4ac
Δ = 102-4·4·(-5)
Δ = 180
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:

$\sqrt{\Delta}=\sqrt{180}=\sqrt{36*5}=\sqrt{36}*\sqrt{5}=6\sqrt{5}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-6\sqrt{5}}{2*4}=\frac{-10-6\sqrt{5}}{8}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+6\sqrt{5}}{2*4}=\frac{-10+6\sqrt{5}}{8}
$