(40-x)(x-5)=4(40-x)

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Solution for (40-x)(x-5)=4(40-x) equation:



(40-x)(x-5)=4(40-x)
We move all terms to the left:
(40-x)(x-5)-(4(40-x))=0
We add all the numbers together, and all the variables
(-1x+40)(x-5)-(4(-1x+40))=0
We multiply parentheses ..
(-1x^2+5x+40x-200)-(4(-1x+40))=0
We calculate terms in parentheses: -(4(-1x+40)), so:
4(-1x+40)
We multiply parentheses
-4x+160
Back to the equation:
-(-4x+160)
We get rid of parentheses
-1x^2+5x+40x+4x-200-160=0
We add all the numbers together, and all the variables
-1x^2+49x-360=0
a = -1; b = 49; c = -360;
Δ = b2-4ac
Δ = 492-4·(-1)·(-360)
Δ = 961
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{961}=31$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(49)-31}{2*-1}=\frac{-80}{-2} =+40 $
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(49)+31}{2*-1}=\frac{-18}{-2} =+9 $

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