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

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

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

We move all terms to the left:

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

We multiply parentheses

-4x-(3-3x(3x+2))-6+1=0


We calculate terms in parentheses: -(3-3x(3x+2)), so:

3-3x(3x+2)

determiningTheFunctionDomain
-3x(3x+2)+3

We multiply parentheses

-9x^2-6x+3

Back to the equation:

-(-9x^2-6x+3)


We add all the numbers together, and all the variables

-(-9x^2-6x+3)-4x-5=0

We get rid of parentheses

9x^2+6x-4x-3-5=0

We add all the numbers together, and all the variables

9x^2+2x-8=0

a = 9; b = 2; c = -8;
Δ = b2-4ac
Δ = 22-4·9·(-8)
Δ = 292
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{292}=\sqrt{4*73}=\sqrt{4}*\sqrt{73}=2\sqrt{73}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2\sqrt{73}}{2*9}=\frac{-2-2\sqrt{73}}{18}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2\sqrt{73}}{2*9}=\frac{-2+2\sqrt{73}}{18}
$