(3x+5)(4x+3)=(5x-3)(2x-9)+80x+20

Solution for (3x+5)(4x+3)=(5x-3)(2x-9)+80x+20 equation:

(3x+5)(4x+3)=(5x-3)(2x-9)+80x+20

We move all terms to the left:

(3x+5)(4x+3)-((5x-3)(2x-9)+80x+20)=0

We multiply parentheses ..

(+12x^2+9x+20x+15)-((5x-3)(2x-9)+80x+20)=0


We calculate terms in parentheses: -((5x-3)(2x-9)+80x+20), so:

(5x-3)(2x-9)+80x+20

We add all the numbers together, and all the variables

80x+(5x-3)(2x-9)+20

We multiply parentheses ..

(+10x^2-45x-6x+27)+80x+20

We get rid of parentheses

10x^2-45x-6x+80x+27+20

We add all the numbers together, and all the variables

10x^2+29x+47

Back to the equation:

-(10x^2+29x+47)


We get rid of parentheses

12x^2-10x^2+9x+20x-29x+15-47=0

We add all the numbers together, and all the variables

2x^2-32=0

a = 2; b = 0; c = -32;
Δ = b2-4ac
Δ = 02-4·2·(-32)
Δ = 256
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{256}=16$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-16}{2*2}=\frac{-16}{4}
=-4
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+16}{2*2}=\frac{16}{4}
=4
$