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9x^2+6x-3=0
a = 9; b = 6; c = -3;
Δ = b2-4ac
Δ = 62-4·9·(-3)
Δ = 144
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{144}=12$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-12}{2*9}=\frac{-18}{18} =-1 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+12}{2*9}=\frac{6}{18} =1/3 $
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