0=(2x-9)*(x+5))/(x-4)

Solution for 0=(2x-9)*(x+5))/(x-4) equation:

0=(2x-9)(x+5))/(x-4)

We move all terms to the left:

0-((2x-9)(x+5))/(x-4))=0


Domain of the equation: (x-4))+0!=0

x∈R


We multiply parentheses ..

-((+2x^2+10x-9x-45))/(x-4))+0=0

We multiply all the terms by the denominator


-((+2x^2+10x-9x-45))=0


We calculate terms in parentheses: -((+2x^2+10x-9x-45)), so:

(+2x^2+10x-9x-45)

We get rid of parentheses

2x^2+10x-9x-45

We add all the numbers together, and all the variables

2x^2+x-45

Back to the equation:

-(2x^2+x-45)


We get rid of parentheses

-2x^2-x+45=0

We add all the numbers together, and all the variables

-2x^2-1x+45=0

a = -2; b = -1; c = +45;
Δ = b2-4ac
Δ = -12-4·(-2)·45
Δ = 361
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{361}=19$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1)-19}{2*-2}=\frac{-18}{-4}
=4+1/2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+19}{2*-2}=\frac{20}{-4}
=-5
$