(x-4)(x+7)+(x-4)(2x-5)=0

Solution for (x-4)(x+7)+(x-4)(2x-5)=0 equation:

(x-4)(x+7)+(x-4)(2x-5)=0

We multiply parentheses ..

(+x^2+7x-4x-28)+(x-4)(2x-5)=0

We get rid of parentheses

x^2+7x-4x+(x-4)(2x-5)-28=0

We multiply parentheses ..

x^2+(+2x^2-5x-8x+20)+7x-4x-28=0

We add all the numbers together, and all the variables

x^2+(+2x^2-5x-8x+20)+3x-28=0

We get rid of parentheses

x^2+2x^2-5x-8x+3x+20-28=0

We add all the numbers together, and all the variables

3x^2-10x-8=0

a = 3; b = -10; c = -8;
Δ = b2-4ac
Δ = -102-4·3·(-8)
Δ = 196
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{196}=14$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-14}{2*3}=\frac{-4}{6}
=-2/3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+14}{2*3}=\frac{24}{6}
=4
$