54=4x+5x(2+2x)+12+2x

Solution for 54=4x+5x(2+2x)+12+2x equation:

54=4x+5x(2+2x)+12+2x

We move all terms to the left:

54-(4x+5x(2+2x)+12+2x)=0

We add all the numbers together, and all the variables

-(4x+5x(2x+2)+12+2x)+54=0


We calculate terms in parentheses: -(4x+5x(2x+2)+12+2x), so:

4x+5x(2x+2)+12+2x

determiningTheFunctionDomain
4x+5x(2x+2)+2x+12

We add all the numbers together, and all the variables

6x+5x(2x+2)+12

We multiply parentheses

10x^2+6x+10x+12

We add all the numbers together, and all the variables

10x^2+16x+12

Back to the equation:

-(10x^2+16x+12)


We get rid of parentheses

-10x^2-16x-12+54=0

We add all the numbers together, and all the variables

-10x^2-16x+42=0

a = -10; b = -16; c = +42;
Δ = b2-4ac
Δ = -162-4·(-10)·42
Δ = 1936
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1936}=44$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-16)-44}{2*-10}=\frac{-28}{-20}
=1+2/5
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16)+44}{2*-10}=\frac{60}{-20}
=-3
$