(3x-4)(x+5)+(3x-4)(5x-1)=0

Solution for (3x-4)(x+5)+(3x-4)(5x-1)=0 equation:

(3x-4)(x+5)+(3x-4)(5x-1)=0

We multiply parentheses ..

(+3x^2+15x-4x-20)+(3x-4)(5x-1)=0

We get rid of parentheses

3x^2+15x-4x+(3x-4)(5x-1)-20=0

We multiply parentheses ..

3x^2+(+15x^2-3x-20x+4)+15x-4x-20=0

We add all the numbers together, and all the variables

3x^2+(+15x^2-3x-20x+4)+11x-20=0

We get rid of parentheses

3x^2+15x^2-3x-20x+11x+4-20=0

We add all the numbers together, and all the variables

18x^2-12x-16=0

a = 18; b = -12; c = -16;
Δ = b2-4ac
Δ = -122-4·18·(-16)
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{-(-12)-36}{2*18}=\frac{-24}{36}
=-2/3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+36}{2*18}=\frac{48}{36}
=1+1/3
$