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4v^2-33v+35=0
a = 4; b = -33; c = +35;
Δ = b2-4ac
Δ = -332-4·4·35
Δ = 529
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{529}=23$$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-33)-23}{2*4}=\frac{10}{8} =1+1/4 $$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-33)+23}{2*4}=\frac{56}{8} =7 $
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