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2d+(2^3-5)=10(5-2)+d(12/6)
We move all terms to the left:
2d+(2^3-5)-(10(5-2)+d(12/6))=0
We add all the numbers together, and all the variables
2d-(103+d2)+(2^3-5)=0
We add all the numbers together, and all the variables
2d-(103+d2)+3=0
We get rid of parentheses
2d-d2-103+3=0
We add all the numbers together, and all the variables
-1d^2+2d-100=0
a = -1; b = 2; c = -100;
Δ = b2-4ac
Δ = 22-4·(-1)·(-100)
Δ = -396
Delta is less than zero, so there is no solution for the equation
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