5x+1=(3-4x)(2+3x)

Solution for 5x+1=(3-4x)(2+3x) equation:

5x+1=(3-4x)(2+3x)

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

-((-12x^2-8x+9x+6))+5x+1=0


We calculate terms in parentheses: -((-12x^2-8x+9x+6)), so:

(-12x^2-8x+9x+6)

We get rid of parentheses

-12x^2-8x+9x+6

We add all the numbers together, and all the variables

-12x^2+x+6

Back to the equation:

-(-12x^2+x+6)


We get rid of parentheses

12x^2-x+5x-6+1=0

We add all the numbers together, and all the variables

12x^2+4x-5=0

a = 12; b = 4; c = -5;
Δ = b2-4ac
Δ = 42-4·12·(-5)
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{-(4)-16}{2*12}=\frac{-20}{24}
=-5/6
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+16}{2*12}=\frac{12}{24}
=1/2
$