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2(d-2)=12+2/3d
We move all terms to the left:
2(d-2)-(12+2/3d)=0
Domain of the equation: 3d)!=0We add all the numbers together, and all the variables
d!=0/1
d!=0
d∈R
2(d-2)-(2/3d+12)=0
We multiply parentheses
2d-(2/3d+12)-4=0
We get rid of parentheses
2d-2/3d-12-4=0
We multiply all the terms by the denominator
2d*3d-12*3d-4*3d-2=0
Wy multiply elements
6d^2-36d-12d-2=0
We add all the numbers together, and all the variables
6d^2-48d-2=0
a = 6; b = -48; c = -2;
Δ = b2-4ac
Δ = -482-4·6·(-2)
Δ = 2352
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$d_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$d_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{2352}=\sqrt{784*3}=\sqrt{784}*\sqrt{3}=28\sqrt{3}$$d_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-48)-28\sqrt{3}}{2*6}=\frac{48-28\sqrt{3}}{12} $$d_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-48)+28\sqrt{3}}{2*6}=\frac{48+28\sqrt{3}}{12} $
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