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3x^2-2(1)x+(1)=0
a = 3; b = -21; c = +1;
Δ = b2-4ac
Δ = -212-4·3·1
Δ = 429
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-21)-\sqrt{429}}{2*3}=\frac{21-\sqrt{429}}{6} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-21)+\sqrt{429}}{2*3}=\frac{21+\sqrt{429}}{6} $
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