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3v^2+17v+24=5
We move all terms to the left:
3v^2+17v+24-(5)=0
We add all the numbers together, and all the variables
3v^2+17v+19=0
a = 3; b = 17; c = +19;
Δ = b2-4ac
Δ = 172-4·3·19
Δ = 61
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}$$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(17)-\sqrt{61}}{2*3}=\frac{-17-\sqrt{61}}{6} $$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(17)+\sqrt{61}}{2*3}=\frac{-17+\sqrt{61}}{6} $
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