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3x^2+3x^2=27x
We move all terms to the left:
3x^2+3x^2-(27x)=0
We add all the numbers together, and all the variables
6x^2-27x=0
a = 6; b = -27; c = 0;
Δ = b2-4ac
Δ = -272-4·6·0
Δ = 729
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{729}=27$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-27)-27}{2*6}=\frac{0}{12} =0 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-27)+27}{2*6}=\frac{54}{12} =4+1/2 $
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