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13x-5x^2=0
a = -5; b = 13; c = 0;
Δ = b2-4ac
Δ = 132-4·(-5)·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$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(13)-13}{2*-5}=\frac{-26}{-10} =2+3/5 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(13)+13}{2*-5}=\frac{0}{-10} =0 $
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