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

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

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

We multiply parentheses ..

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

We get rid of parentheses

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

We multiply parentheses ..

x^2+(+3x^2-4x+9x-12)-5x+3x-15=0

We add all the numbers together, and all the variables

x^2+(+3x^2-4x+9x-12)-2x-15=0

We get rid of parentheses

x^2+3x^2-4x+9x-2x-12-15=0

We add all the numbers together, and all the variables

4x^2+3x-27=0

a = 4; b = 3; c = -27;
Δ = b2-4ac
Δ = 32-4·4·(-27)
Δ = 441
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{441}=21$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-21}{2*4}=\frac{-24}{8}
=-3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+21}{2*4}=\frac{18}{8}
=2+1/4
$