(9x+18)(x+2)=(4x+20)(x+5)

Solution for (9x+18)(x+2)=(4x+20)(x+5) equation:

(9x+18)(x+2)=(4x+20)(x+5)

We move all terms to the left:

(9x+18)(x+2)-((4x+20)(x+5))=0

We multiply parentheses ..

(+9x^2+18x+18x+36)-((4x+20)(x+5))=0


We calculate terms in parentheses: -((4x+20)(x+5)), so:

(4x+20)(x+5)

We multiply parentheses ..

(+4x^2+20x+20x+100)

We get rid of parentheses

4x^2+20x+20x+100

We add all the numbers together, and all the variables

4x^2+40x+100

Back to the equation:

-(4x^2+40x+100)


We get rid of parentheses

9x^2-4x^2+18x+18x-40x+36-100=0

We add all the numbers together, and all the variables

5x^2-4x-64=0

a = 5; b = -4; c = -64;
Δ = b2-4ac
Δ = -42-4·5·(-64)
Δ = 1296
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1296}=36$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-36}{2*5}=\frac{-32}{10}
=-3+1/5
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+36}{2*5}=\frac{40}{10}
=4
$