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50x^2+62x=0
a = 50; b = 62; c = 0;
Δ = b2-4ac
Δ = 622-4·50·0
Δ = 3844
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{3844}=62$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(62)-62}{2*50}=\frac{-124}{100} =-1+6/25 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(62)+62}{2*50}=\frac{0}{100} =0 $
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