y=1+0.01y^2

Solution for y=1+0.01y^2 equation:

y=1+0.01y^2

We move all terms to the left:

y-(1+0.01y^2)=0

We get rid of parentheses

-0.01y^2+y-1=0

a = -0.01; b = 1; c = -1;
Δ = b2-4ac
Δ = 12-4·(-0.01)·(-1)
Δ = 0.96
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{0.96}}{2*-0.01}=\frac{-1-\sqrt{0.96}}{-0.02}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{0.96}}{2*-0.01}=\frac{-1+\sqrt{0.96}}{-0.02}
$