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2n^2-13n=0
a = 2; b = -13; c = 0;
Δ = b2-4ac
Δ = -132-4·2·0
Δ = 169
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{169}=13$$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-13)-13}{2*2}=\frac{0}{4} =0 $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-13)+13}{2*2}=\frac{26}{4} =6+1/2 $
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