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36x^2+13x=0
a = 36; b = 13; c = 0;
Δ = b2-4ac
Δ = 132-4·36·0
Δ = 169
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{169}=13$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(13)-13}{2*36}=\frac{-26}{72} =-13/36 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(13)+13}{2*36}=\frac{0}{72} =0 $
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