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9x^2+18x+9=49
We move all terms to the left:
9x^2+18x+9-(49)=0
We add all the numbers together, and all the variables
9x^2+18x-40=0
a = 9; b = 18; c = -40;
Δ = b2-4ac
Δ = 182-4·9·(-40)
Δ = 1764
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{1764}=42$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(18)-42}{2*9}=\frac{-60}{18} =-3+1/3 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(18)+42}{2*9}=\frac{24}{18} =1+1/3 $
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