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28v^2-14v=0
a = 28; b = -14; c = 0;
Δ = b2-4ac
Δ = -142-4·28·0
Δ = 196
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{196}=14$$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-14)-14}{2*28}=\frac{0}{56} =0 $$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-14)+14}{2*28}=\frac{28}{56} =1/2 $
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