44-(2x+3)=4(x+5)x

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Solution for 44-(2x+3)=4(x+5)x equation:



44-(2x+3)=4(x+5)x
We move all terms to the left:
44-(2x+3)-(4(x+5)x)=0
We get rid of parentheses
-2x-(4(x+5)x)-3+44=0
We calculate terms in parentheses: -(4(x+5)x), so:
4(x+5)x
We multiply parentheses
4x^2+20x
Back to the equation:
-(4x^2+20x)
We add all the numbers together, and all the variables
-2x-(4x^2+20x)+41=0
We get rid of parentheses
-4x^2-2x-20x+41=0
We add all the numbers together, and all the variables
-4x^2-22x+41=0
a = -4; b = -22; c = +41;
Δ = b2-4ac
Δ = -222-4·(-4)·41
Δ = 1140
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:
$\sqrt{\Delta}=\sqrt{1140}=\sqrt{4*285}=\sqrt{4}*\sqrt{285}=2\sqrt{285}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-22)-2\sqrt{285}}{2*-4}=\frac{22-2\sqrt{285}}{-8} $
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-22)+2\sqrt{285}}{2*-4}=\frac{22+2\sqrt{285}}{-8} $

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