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-3v^2+13v-12=0
a = -3; b = 13; c = -12;
Δ = b2-4ac
Δ = 132-4·(-3)·(-12)
Δ = 25
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{25}=5$$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(13)-5}{2*-3}=\frac{-18}{-6} =+3 $$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(13)+5}{2*-3}=\frac{-8}{-6} =1+1/3 $
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