4y^2+22y-10=0

Solution for 4y^2+22y-10=0 equation:

4y^2+22y-10=0

a = 4; b = 22; c = -10;
Δ = b2-4ac
Δ = 222-4·4·(-10)
Δ = 644
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:

$\sqrt{\Delta}=\sqrt{644}=\sqrt{4*161}=\sqrt{4}*\sqrt{161}=2\sqrt{161}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(22)-2\sqrt{161}}{2*4}=\frac{-22-2\sqrt{161}}{8}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(22)+2\sqrt{161}}{2*4}=\frac{-22+2\sqrt{161}}{8}
$