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15v^2-65v+60=0
a = 15; b = -65; c = +60;
Δ = b2-4ac
Δ = -652-4·15·60
Δ = 625
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{625}=25$$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-65)-25}{2*15}=\frac{40}{30} =1+1/3 $$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-65)+25}{2*15}=\frac{90}{30} =3 $
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