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3b^2-b=5
We move all terms to the left:
3b^2-b-(5)=0
We add all the numbers together, and all the variables
3b^2-1b-5=0
a = 3; b = -1; c = -5;
Δ = b2-4ac
Δ = -12-4·3·(-5)
Δ = 61
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1)-\sqrt{61}}{2*3}=\frac{1-\sqrt{61}}{6} $$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+\sqrt{61}}{2*3}=\frac{1+\sqrt{61}}{6} $
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