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5v^2-13v-28=0
a = 5; b = -13; c = -28;
Δ = b2-4ac
Δ = -132-4·5·(-28)
Δ = 729
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{729}=27$$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-13)-27}{2*5}=\frac{-14}{10} =-1+2/5 $$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-13)+27}{2*5}=\frac{40}{10} =4 $
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