(x+3)(4x-5)=(x+3)(3x-11)

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Solution for (x+3)(4x-5)=(x+3)(3x-11) equation:



(x+3)(4x-5)=(x+3)(3x-11)
We move all terms to the left:
(x+3)(4x-5)-((x+3)(3x-11))=0
We multiply parentheses ..
(+4x^2-5x+12x-15)-((x+3)(3x-11))=0
We calculate terms in parentheses: -((x+3)(3x-11)), so:
(x+3)(3x-11)
We multiply parentheses ..
(+3x^2-11x+9x-33)
We get rid of parentheses
3x^2-11x+9x-33
We add all the numbers together, and all the variables
3x^2-2x-33
Back to the equation:
-(3x^2-2x-33)
We get rid of parentheses
4x^2-3x^2-5x+12x+2x-15+33=0
We add all the numbers together, and all the variables
x^2+9x+18=0
a = 1; b = 9; c = +18;
Δ = b2-4ac
Δ = 92-4·1·18
Δ = 9
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{9}=3$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(9)-3}{2*1}=\frac{-12}{2} =-6 $
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(9)+3}{2*1}=\frac{-6}{2} =-3 $

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