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3x+9x^2=12
We move all terms to the left:
3x+9x^2-(12)=0
a = 9; b = 3; c = -12;
Δ = b2-4ac
Δ = 32-4·9·(-12)
Δ = 441
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{441}=21$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-21}{2*9}=\frac{-24}{18} =-1+1/3 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+21}{2*9}=\frac{18}{18} =1 $
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