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

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

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

We multiply parentheses ..

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

We get rid of parentheses

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

We multiply parentheses ..

x^2+(+4x^2-5x-36x+45)+4x-9x-36=0

We add all the numbers together, and all the variables

x^2+(+4x^2-5x-36x+45)-5x-36=0

We get rid of parentheses

x^2+4x^2-5x-36x-5x+45-36=0

We add all the numbers together, and all the variables

5x^2-46x+9=0

a = 5; b = -46; c = +9;
Δ = b2-4ac
Δ = -462-4·5·9
Δ = 1936
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1936}=44$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-46)-44}{2*5}=\frac{2}{10}
=1/5
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-46)+44}{2*5}=\frac{90}{10}
=9
$