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9x^2+3x+4=9
We move all terms to the left:
9x^2+3x+4-(9)=0
We add all the numbers together, and all the variables
9x^2+3x-5=0
a = 9; b = 3; c = -5;
Δ = b2-4ac
Δ = 32-4·9·(-5)
Δ = 189
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{189}=\sqrt{9*21}=\sqrt{9}*\sqrt{21}=3\sqrt{21}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-3\sqrt{21}}{2*9}=\frac{-3-3\sqrt{21}}{18} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+3\sqrt{21}}{2*9}=\frac{-3+3\sqrt{21}}{18} $
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