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8v=4v^2-4
We move all terms to the left:
8v-(4v^2-4)=0
We get rid of parentheses
-4v^2+8v+4=0
a = -4; b = 8; c = +4;
Δ = b2-4ac
Δ = 82-4·(-4)·4
Δ = 128
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}$$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{128}=\sqrt{64*2}=\sqrt{64}*\sqrt{2}=8\sqrt{2}$$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-8\sqrt{2}}{2*-4}=\frac{-8-8\sqrt{2}}{-8} $$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+8\sqrt{2}}{2*-4}=\frac{-8+8\sqrt{2}}{-8} $
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