13x=((4x+x)(4x+x))-3x-((4-x)(4-x))

Solution for 13x=((4x+x)(4x+x))-3x-((4-x)(4-x)) equation:

13x=((4x+x)(4x+x))-3x-((4-x)(4-x))

We move all terms to the left:

13x-(((4x+x)(4x+x))-3x-((4-x)(4-x)))=0

We add all the numbers together, and all the variables

13x-(((+5x)(+5x))-3x-((-1x+4)(-1x+4)))=0

We multiply parentheses ..

-(((+25x^2))-3x-((-1x+4)(-1x+4)))+13x=0


We calculate terms in parentheses: -(((+25x^2))-3x-((-1x+4)(-1x+4))), so:

((+25x^2))-3x-((-1x+4)(-1x+4))

We multiply parentheses ..

((+25x^2))-((+x^2-4x-4x+16))-3x


We calculate terms in parentheses: +((+25x^2)), so:

(+25x^2)

We get rid of parentheses

25x^2

Back to the equation:

+(25x^2)



We calculate terms in parentheses: -((+x^2-4x-4x+16)), so:

(+x^2-4x-4x+16)

We get rid of parentheses

x^2-4x-4x+16

We add all the numbers together, and all the variables

x^2-8x+16

Back to the equation:

-(x^2-8x+16)


We add all the numbers together, and all the variables

25x^2-3x-(x^2-8x+16)

We get rid of parentheses

25x^2-x^2-3x+8x-16

We add all the numbers together, and all the variables

24x^2+5x-16

Back to the equation:

-(24x^2+5x-16)


We add all the numbers together, and all the variables

13x-(24x^2+5x-16)=0

We get rid of parentheses

-24x^2+13x-5x+16=0

We add all the numbers together, and all the variables

-24x^2+8x+16=0

a = -24; b = 8; c = +16;
Δ = b2-4ac
Δ = 82-4·(-24)·16
Δ = 1600
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1600}=40$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-40}{2*-24}=\frac{-48}{-48}
=1
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+40}{2*-24}=\frac{32}{-48}
=-2/3
$