(2x-3)(x+18)=(x-6)(x+9)

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Solution for (2x-3)(x+18)=(x-6)(x+9) equation:



(2x-3)(x+18)=(x-6)(x+9)
We move all terms to the left:
(2x-3)(x+18)-((x-6)(x+9))=0
We multiply parentheses ..
(+2x^2+36x-3x-54)-((x-6)(x+9))=0
We calculate terms in parentheses: -((x-6)(x+9)), so:
(x-6)(x+9)
We multiply parentheses ..
(+x^2+9x-6x-54)
We get rid of parentheses
x^2+9x-6x-54
We add all the numbers together, and all the variables
x^2+3x-54
Back to the equation:
-(x^2+3x-54)
We get rid of parentheses
2x^2-x^2+36x-3x-3x-54+54=0
We add all the numbers together, and all the variables
x^2+30x=0
a = 1; b = 30; c = 0;
Δ = b2-4ac
Δ = 302-4·1·0
Δ = 900
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{900}=30$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(30)-30}{2*1}=\frac{-60}{2} =-30 $
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(30)+30}{2*1}=\frac{0}{2} =0 $

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