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4v^2+2v-1=5
We move all terms to the left:
4v^2+2v-1-(5)=0
We add all the numbers together, and all the variables
4v^2+2v-6=0
a = 4; b = 2; c = -6;
Δ = b2-4ac
Δ = 22-4·4·(-6)
Δ = 100
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{100}=10$$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-10}{2*4}=\frac{-12}{8} =-1+1/2 $$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+10}{2*4}=\frac{8}{8} =1 $
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