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3x+6x^2=9
We move all terms to the left:
3x+6x^2-(9)=0
a = 6; b = 3; c = -9;
Δ = b2-4ac
Δ = 32-4·6·(-9)
Δ = 225
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{225}=15$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-15}{2*6}=\frac{-18}{12} =-1+1/2 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+15}{2*6}=\frac{12}{12} =1 $
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