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4v^2-15v+16=6
We move all terms to the left:
4v^2-15v+16-(6)=0
We add all the numbers together, and all the variables
4v^2-15v+10=0
a = 4; b = -15; c = +10;
Δ = b2-4ac
Δ = -152-4·4·10
Δ = 65
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}$$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-15)-\sqrt{65}}{2*4}=\frac{15-\sqrt{65}}{8} $$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-15)+\sqrt{65}}{2*4}=\frac{15+\sqrt{65}}{8} $
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