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

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
$