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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} $
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