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29x^2-25x=0
a = 29; b = -25; c = 0;
Δ = b2-4ac
Δ = -252-4·29·0
Δ = 625
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{625}=25$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-25)-25}{2*29}=\frac{0}{58} =0 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-25)+25}{2*29}=\frac{50}{58} =25/29 $
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