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14x^2-28x=0
a = 14; b = -28; c = 0;
Δ = b2-4ac
Δ = -282-4·14·0
Δ = 784
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{784}=28$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-28)-28}{2*14}=\frac{0}{28} =0 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-28)+28}{2*14}=\frac{56}{28} =2 $
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