2y+y^2=7

Solution for 2y+y^2=7 equation:

2y+y^2=7

We move all terms to the left:

2y+y^2-(7)=0

a = 1; b = 2; c = -7;
Δ = b2-4ac
Δ = 22-4·1·(-7)
Δ = 32
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:

$\sqrt{\Delta}=\sqrt{32}=\sqrt{16*2}=\sqrt{16}*\sqrt{2}=4\sqrt{2}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-4\sqrt{2}}{2*1}=\frac{-2-4\sqrt{2}}{2}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+4\sqrt{2}}{2*1}=\frac{-2+4\sqrt{2}}{2}
$