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x(3-1)+4(3x+4)=18-6x
We move all terms to the left:
x(3-1)+4(3x+4)-(18-6x)=0
We add all the numbers together, and all the variables
x2+4(3x+4)-(-6x+18)=0
We add all the numbers together, and all the variables
x^2+4(3x+4)-(-6x+18)=0
We multiply parentheses
x^2+12x-(-6x+18)+16=0
We get rid of parentheses
x^2+12x+6x-18+16=0
We add all the numbers together, and all the variables
x^2+18x-2=0
a = 1; b = 18; c = -2;
Δ = b2-4ac
Δ = 182-4·1·(-2)
Δ = 332
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{332}=\sqrt{4*83}=\sqrt{4}*\sqrt{83}=2\sqrt{83}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(18)-2\sqrt{83}}{2*1}=\frac{-18-2\sqrt{83}}{2} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(18)+2\sqrt{83}}{2*1}=\frac{-18+2\sqrt{83}}{2} $
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