y^2+10y=3

Solution for y^2+10y=3 equation:

y^2+10y=3

We move all terms to the left:

y^2+10y-(3)=0

a = 1; b = 10; c = -3;
Δ = b2-4ac
Δ = 102-4·1·(-3)
Δ = 112
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:

$\sqrt{\Delta}=\sqrt{112}=\sqrt{16*7}=\sqrt{16}*\sqrt{7}=4\sqrt{7}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-4\sqrt{7}}{2*1}=\frac{-10-4\sqrt{7}}{2}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+4\sqrt{7}}{2*1}=\frac{-10+4\sqrt{7}}{2}
$