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2v^2-32=-12v
We move all terms to the left:
2v^2-32-(-12v)=0
We get rid of parentheses
2v^2+12v-32=0
a = 2; b = 12; c = -32;
Δ = b2-4ac
Δ = 122-4·2·(-32)
Δ = 400
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{400}=20$$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-20}{2*2}=\frac{-32}{4} =-8 $$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+20}{2*2}=\frac{8}{4} =2 $
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