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3d^2-3=d
We move all terms to the left:
3d^2-3-(d)=0
We add all the numbers together, and all the variables
3d^2-1d-3=0
a = 3; b = -1; c = -3;
Δ = b2-4ac
Δ = -12-4·3·(-3)
Δ = 37
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}$$d_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$d_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1)-\sqrt{37}}{2*3}=\frac{1-\sqrt{37}}{6} $$d_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+\sqrt{37}}{2*3}=\frac{1+\sqrt{37}}{6} $
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