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-3x^2+13x-9=0
a = -3; b = 13; c = -9;
Δ = b2-4ac
Δ = 132-4·(-3)·(-9)
Δ = 61
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{-(13)-\sqrt{61}}{2*-3}=\frac{-13-\sqrt{61}}{-6} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(13)+\sqrt{61}}{2*-3}=\frac{-13+\sqrt{61}}{-6} $
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