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3a^2=a+6
We move all terms to the left:
3a^2-(a+6)=0
We get rid of parentheses
3a^2-a-6=0
We add all the numbers together, and all the variables
3a^2-1a-6=0
a = 3; b = -1; c = -6;
Δ = b2-4ac
Δ = -12-4·3·(-6)
Δ = 73
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1)-\sqrt{73}}{2*3}=\frac{1-\sqrt{73}}{6} $$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+\sqrt{73}}{2*3}=\frac{1+\sqrt{73}}{6} $
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