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36y^2-13y+1=0
a = 36; b = -13; c = +1;
Δ = b2-4ac
Δ = -132-4·36·1
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$$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-13)-5}{2*36}=\frac{8}{72} =1/9 $$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-13)+5}{2*36}=\frac{18}{72} =1/4 $
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