1+3x(x-2)=36-(X+6)

Solution for 1+3x(x-2)=36-(X+6) equation:

1+3x(x-2)=36-(x+6)

We move all terms to the left:

1+3x(x-2)-(36-(x+6))=0

We multiply parentheses

3x^2-6x-(36-(x+6))+1=0


We calculate terms in parentheses: -(36-(x+6)), so:

36-(x+6)

determiningTheFunctionDomain
-(x+6)+36

We get rid of parentheses

-x-6+36

We add all the numbers together, and all the variables

-1x+30

Back to the equation:

-(-1x+30)


We get rid of parentheses

3x^2-6x+1x-30+1=0

We add all the numbers together, and all the variables

3x^2-5x-29=0

a = 3; b = -5; c = -29;
Δ = b2-4ac
Δ = -52-4·3·(-29)
Δ = 373
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-5)-\sqrt{373}}{2*3}=\frac{5-\sqrt{373}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+\sqrt{373}}{2*3}=\frac{5+\sqrt{373}}{6}
$