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8+2v-v^2=0
We add all the numbers together, and all the variables
-1v^2+2v+8=0
a = -1; b = 2; c = +8;
Δ = b2-4ac
Δ = 22-4·(-1)·8
Δ = 36
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{36}=6$$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-6}{2*-1}=\frac{-8}{-2} =+4 $$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+6}{2*-1}=\frac{4}{-2} =-2 $
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