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13x^2+14x=0
a = 13; b = 14; c = 0;
Δ = b2-4ac
Δ = 142-4·13·0
Δ = 196
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{196}=14$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(14)-14}{2*13}=\frac{-28}{26} =-1+1/13 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(14)+14}{2*13}=\frac{0}{26} =0 $
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