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3(x+2)(x-2)=(x-4)+8x
We move all terms to the left:
3(x+2)(x-2)-((x-4)+8x)=0
We use the square of the difference formula
x^2-((x-4)+8x)-4=0
We calculate terms in parentheses: -((x-4)+8x), so:We get rid of parentheses
(x-4)+8x
We add all the numbers together, and all the variables
8x+(x-4)
We get rid of parentheses
8x+x-4
We add all the numbers together, and all the variables
9x-4
Back to the equation:
-(9x-4)
x^2-9x+4-4=0
We add all the numbers together, and all the variables
x^2-9x=0
a = 1; b = -9; c = 0;
Δ = b2-4ac
Δ = -92-4·1·0
Δ = 81
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{81}=9$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-9)-9}{2*1}=\frac{0}{2} =0 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-9)+9}{2*1}=\frac{18}{2} =9 $
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