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(2x-16)(6+2x-18)=(11-x)(x-3)
We move all terms to the left:
(2x-16)(6+2x-18)-((11-x)(x-3))=0
We add all the numbers together, and all the variables
(2x-16)(2x-12)-((-1x+11)(x-3))=0
We multiply parentheses ..
(+4x^2-24x-32x+192)-((-1x+11)(x-3))=0
We calculate terms in parentheses: -((-1x+11)(x-3)), so:We get rid of parentheses
(-1x+11)(x-3)
We multiply parentheses ..
(-1x^2+3x+11x-33)
We get rid of parentheses
-1x^2+3x+11x-33
We add all the numbers together, and all the variables
-1x^2+14x-33
Back to the equation:
-(-1x^2+14x-33)
4x^2+1x^2-24x-32x-14x+192+33=0
We add all the numbers together, and all the variables
5x^2-70x+225=0
a = 5; b = -70; c = +225;
Δ = b2-4ac
Δ = -702-4·5·225
Δ = 400
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{400}=20$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-70)-20}{2*5}=\frac{50}{10} =5 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-70)+20}{2*5}=\frac{90}{10} =9 $
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