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(3x)^2-10(3x)+9=0
a = 3; b = -103; c = +9;
Δ = b2-4ac
Δ = -1032-4·3·9
Δ = 10501
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{-(-103)-\sqrt{10501}}{2*3}=\frac{103-\sqrt{10501}}{6} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-103)+\sqrt{10501}}{2*3}=\frac{103+\sqrt{10501}}{6} $
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