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5v^2+23v-10=0
a = 5; b = 23; c = -10;
Δ = b2-4ac
Δ = 232-4·5·(-10)
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$$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(23)-27}{2*5}=\frac{-50}{10} =-5 $$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(23)+27}{2*5}=\frac{4}{10} =2/5 $
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