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6d^2+21d=10d
We move all terms to the left:
6d^2+21d-(10d)=0
We add all the numbers together, and all the variables
6d^2+11d=0
a = 6; b = 11; c = 0;
Δ = b2-4ac
Δ = 112-4·6·0
Δ = 121
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}$$d_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{121}=11$$d_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(11)-11}{2*6}=\frac{-22}{12} =-1+5/6 $$d_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(11)+11}{2*6}=\frac{0}{12} =0 $
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