5y^2+212y+22=0

Solution for 5y^2+212y+22=0 equation:

5y^2+212y+22=0

a = 5; b = 212; c = +22;
Δ = b2-4ac
Δ = 2122-4·5·22
Δ = 44504
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:

$\sqrt{\Delta}=\sqrt{44504}=\sqrt{4*11126}=\sqrt{4}*\sqrt{11126}=2\sqrt{11126}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(212)-2\sqrt{11126}}{2*5}=\frac{-212-2\sqrt{11126}}{10}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(212)+2\sqrt{11126}}{2*5}=\frac{-212+2\sqrt{11126}}{10}
$