y2+y+1=14

Solution for y2+y+1=14 equation:

y2+y+1=14

We move all terms to the left:

y2+y+1-(14)=0

We add all the numbers together, and all the variables

y^2+y-13=0

a = 1; b = 1; c = -13;
Δ = b2-4ac
Δ = 12-4·1·(-13)
Δ = 53
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-\sqrt{53}}{2*1}=\frac{-1-\sqrt{53}}{2}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{53}}{2*1}=\frac{-1+\sqrt{53}}{2}
$