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3v^2+13v+4=0
a = 3; b = 13; c = +4;
Δ = b2-4ac
Δ = 132-4·3·4
Δ = 121
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{121}=11$$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(13)-11}{2*3}=\frac{-24}{6} =-4 $$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(13)+11}{2*3}=\frac{-2}{6} =-1/3 $
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