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3x^2-13x+14=0
a = 3; b = -13; c = +14;
Δ = b2-4ac
Δ = -132-4·3·14
Δ = 1
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{1}=1$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-13)-1}{2*3}=\frac{12}{6} =2 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-13)+1}{2*3}=\frac{14}{6} =2+1/3 $
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