(5x+3)(4x+9)+(5x+3)(-7x-4)=0

Solution for (5x+3)(4x+9)+(5x+3)(-7x-4)=0 equation:

(5x+3)(4x+9)+(5x+3)(-7x-4)=0

We multiply parentheses ..

(+20x^2+45x+12x+27)+(5x+3)(-7x-4)=0

We get rid of parentheses

20x^2+45x+12x+(5x+3)(-7x-4)+27=0

We multiply parentheses ..

20x^2+(-35x^2-20x-21x-12)+45x+12x+27=0

We add all the numbers together, and all the variables

20x^2+(-35x^2-20x-21x-12)+57x+27=0

We get rid of parentheses

20x^2-35x^2-20x-21x+57x-12+27=0

We add all the numbers together, and all the variables

-15x^2+16x+15=0

a = -15; b = 16; c = +15;
Δ = b2-4ac
Δ = 162-4·(-15)·15
Δ = 1156
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1156}=34$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-34}{2*-15}=\frac{-50}{-30}
=1+2/3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+34}{2*-15}=\frac{18}{-30}
=-3/5
$