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3x^2=9-9x
We move all terms to the left:
3x^2-(9-9x)=0
We add all the numbers together, and all the variables
3x^2-(-9x+9)=0
We get rid of parentheses
3x^2+9x-9=0
a = 3; b = 9; c = -9;
Δ = b2-4ac
Δ = 92-4·3·(-9)
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{-(9)-3\sqrt{21}}{2*3}=\frac{-9-3\sqrt{21}}{6} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(9)+3\sqrt{21}}{2*3}=\frac{-9+3\sqrt{21}}{6} $
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