1+y(y=3)

Solution for 1+y(y=3) equation:

1+y(y=3)

We move all terms to the left:

1+y(y-(3))=0

We multiply parentheses

y^2-3y+1=0

a = 1; b = -3; c = +1;
Δ = b2-4ac
Δ = -32-4·1·1
Δ = 5
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{-(-3)-\sqrt{5}}{2*1}=\frac{3-\sqrt{5}}{2}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+\sqrt{5}}{2*1}=\frac{3+\sqrt{5}}{2}
$