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4v^2-12v+5=0
a = 4; b = -12; c = +5;
Δ = b2-4ac
Δ = -122-4·4·5
Δ = 64
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{64}=8$$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-8}{2*4}=\frac{4}{8} =1/2 $$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+8}{2*4}=\frac{20}{8} =2+1/2 $
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