y^2+12y-10=0

Solution for y^2+12y-10=0 equation:

y^2+12y-10=0

a = 1; b = 12; c = -10;
Δ = b2-4ac
Δ = 122-4·1·(-10)
Δ = 184
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:

$\sqrt{\Delta}=\sqrt{184}=\sqrt{4*46}=\sqrt{4}*\sqrt{46}=2\sqrt{46}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-2\sqrt{46}}{2*1}=\frac{-12-2\sqrt{46}}{2}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+2\sqrt{46}}{2*1}=\frac{-12+2\sqrt{46}}{2}
$