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-3a^2-2a+1=0
a = -3; b = -2; c = +1;
Δ = b2-4ac
Δ = -22-4·(-3)·1
Δ = 16
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{16}=4$$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-4}{2*-3}=\frac{-2}{-6} =1/3 $$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+4}{2*-3}=\frac{6}{-6} =-1 $
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