(3x-5)x=(x-6)(x-7)

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Solution for (3x-5)x=(x-6)(x-7) equation:



(3x-5)x=(x-6)(x-7)
We move all terms to the left:
(3x-5)x-((x-6)(x-7))=0
We multiply parentheses
3x^2-5x-((x-6)(x-7))=0
We multiply parentheses ..
3x^2-((+x^2-7x-6x+42))-5x=0
We calculate terms in parentheses: -((+x^2-7x-6x+42)), so:
(+x^2-7x-6x+42)
We get rid of parentheses
x^2-7x-6x+42
We add all the numbers together, and all the variables
x^2-13x+42
Back to the equation:
-(x^2-13x+42)
We add all the numbers together, and all the variables
3x^2-5x-(x^2-13x+42)=0
We get rid of parentheses
3x^2-x^2-5x+13x-42=0
We add all the numbers together, and all the variables
2x^2+8x-42=0
a = 2; b = 8; c = -42;
Δ = b2-4ac
Δ = 82-4·2·(-42)
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{-(8)-20}{2*2}=\frac{-28}{4} =-7 $
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+20}{2*2}=\frac{12}{4} =3 $

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