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2v^2+4v=6
We move all terms to the left:
2v^2+4v-(6)=0
a = 2; b = 4; c = -6;
Δ = b2-4ac
Δ = 42-4·2·(-6)
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$$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-8}{2*2}=\frac{-12}{4} =-3 $$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+8}{2*2}=\frac{4}{4} =1 $
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