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3v^2-5=2v
We move all terms to the left:
3v^2-5-(2v)=0
a = 3; b = -2; c = -5;
Δ = b2-4ac
Δ = -22-4·3·(-5)
Δ = 64
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{64}=8$$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-8}{2*3}=\frac{-6}{6} =-1 $$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+8}{2*3}=\frac{10}{6} =1+2/3 $
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