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6a^2+1-5a=0
a = 6; b = -5; c = +1;
Δ = b2-4ac
Δ = -52-4·6·1
Δ = 1
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}$$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{1}=1$$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-5)-1}{2*6}=\frac{4}{12} =1/3 $$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+1}{2*6}=\frac{6}{12} =1/2 $
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