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10(v+1)-4v=3(*2v+4)-13
We move all terms to the left:
10(v+1)-4v-(3(*2v+4)-13)=0
We add all the numbers together, and all the variables
-4v+10(v+1)-(3(*2v+4)-13)=0
We multiply parentheses
-4v+10v-(3(*2v+4)-13)+10=0
We calculate terms in parentheses: -(3(*2v+4)-13), so:We add all the numbers together, and all the variables
3(*2v+4)-13
We multiply parentheses
6v^2+12-13
We add all the numbers together, and all the variables
6v^2-1
Back to the equation:
-(6v^2-1)
6v-(6v^2-1)+10=0
We get rid of parentheses
-6v^2+6v+1+10=0
We add all the numbers together, and all the variables
-6v^2+6v+11=0
a = -6; b = 6; c = +11;
Δ = b2-4ac
Δ = 62-4·(-6)·11
Δ = 300
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}$$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{300}=\sqrt{100*3}=\sqrt{100}*\sqrt{3}=10\sqrt{3}$$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-10\sqrt{3}}{2*-6}=\frac{-6-10\sqrt{3}}{-12} $$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+10\sqrt{3}}{2*-6}=\frac{-6+10\sqrt{3}}{-12} $
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