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3x^2+26x+168=2x^2
We move all terms to the left:
3x^2+26x+168-(2x^2)=0
determiningTheFunctionDomain 3x^2-2x^2+26x+168=0
We add all the numbers together, and all the variables
x^2+26x+168=0
a = 1; b = 26; c = +168;
Δ = b2-4ac
Δ = 262-4·1·168
Δ = 4
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{4}=2$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(26)-2}{2*1}=\frac{-28}{2} =-14 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(26)+2}{2*1}=\frac{-24}{2} =-12 $
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