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2v^2=64
We move all terms to the left:
2v^2-(64)=0
a = 2; b = 0; c = -64;
Δ = b2-4ac
Δ = 02-4·2·(-64)
Δ = 512
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}$$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{512}=\sqrt{256*2}=\sqrt{256}*\sqrt{2}=16\sqrt{2}$$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-16\sqrt{2}}{2*2}=\frac{0-16\sqrt{2}}{4} =-\frac{16\sqrt{2}}{4} =-4\sqrt{2} $$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+16\sqrt{2}}{2*2}=\frac{0+16\sqrt{2}}{4} =\frac{16\sqrt{2}}{4} =4\sqrt{2} $
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