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3x+6x+2x^2=108
We move all terms to the left:
3x+6x+2x^2-(108)=0
We add all the numbers together, and all the variables
2x^2+9x-108=0
a = 2; b = 9; c = -108;
Δ = b2-4ac
Δ = 92-4·2·(-108)
Δ = 945
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{945}=\sqrt{9*105}=\sqrt{9}*\sqrt{105}=3\sqrt{105}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(9)-3\sqrt{105}}{2*2}=\frac{-9-3\sqrt{105}}{4} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(9)+3\sqrt{105}}{2*2}=\frac{-9+3\sqrt{105}}{4} $
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