y+(y^2+1)=7

Solution for y+(y^2+1)=7 equation:

y+(y^2+1)=7

We move all terms to the left:

y+(y^2+1)-(7)=0

We get rid of parentheses

y^2+y+1-7=0

We add all the numbers together, and all the variables

y^2+y-6=0

a = 1; b = 1; c = -6;
Δ = b2-4ac
Δ = 12-4·1·(-6)
Δ = 25
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{25}=5$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-5}{2*1}=\frac{-6}{2}
=-3
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+5}{2*1}=\frac{4}{2}
=2
$