2x(3-x)-4(x+2)=3x(2-3x)+5+7x^-12

Solution for 2x(3-x)-4(x+2)=3x(2-3x)+5+7x^-12 equation:

2x(3-x)-4(x+2)=3x(2-3x)+5+7x^-12

We move all terms to the left:

2x(3-x)-4(x+2)-(3x(2-3x)+5+7x^-12)=0

We add all the numbers together, and all the variables

2x(-1x+3)-4(x+2)-(3x(-3x+2)+5+7x^-12)=0

We multiply parentheses

-2x^2+6x-4x-(3x(-3x+2)+5+7x^-12)-8=0


We calculate terms in parentheses: -(3x(-3x+2)+5+7x^-12), so:

3x(-3x+2)+5+7x^-12

determiningTheFunctionDomain
3x(-3x+2)+7x^+5-12

We add all the numbers together, and all the variables

7x+3x(-3x+2)-7

We multiply parentheses

-9x^2+7x+6x-7

We add all the numbers together, and all the variables

-9x^2+13x-7

Back to the equation:

-(-9x^2+13x-7)


We add all the numbers together, and all the variables

-2x^2-(-9x^2+13x-7)+2x-8=0

We get rid of parentheses

-2x^2+9x^2-13x+2x+7-8=0

We add all the numbers together, and all the variables

7x^2-11x-1=0

a = 7; b = -11; c = -1;
Δ = b2-4ac
Δ = -112-4·7·(-1)
Δ = 149
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{-(-11)-\sqrt{149}}{2*7}=\frac{11-\sqrt{149}}{14}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-11)+\sqrt{149}}{2*7}=\frac{11+\sqrt{149}}{14}
$