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

Solution for (x+2=)^2-(2x+1)(4-3x) equation:

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

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

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


We calculate terms in parentheses: +(x+2-()^2-(-6x^2+8x-3x+4)), so:

x+2-()^2-(-6x^2+8x-3x+4)

determiningTheFunctionDomain
-(-6x^2+8x-3x+4)+x+2-()^2

We add all the numbers together, and all the variables

-(-6x^2+8x-3x+4)+x

We get rid of parentheses

6x^2-8x+3x+x-4

We add all the numbers together, and all the variables

6x^2-4x-4

Back to the equation:

+(6x^2-4x-4)


We get rid of parentheses

6x^2-4x-4=0

a = 6; b = -4; c = -4;
Δ = b2-4ac
Δ = -42-4·6·(-4)
Δ = 112
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:

$\sqrt{\Delta}=\sqrt{112}=\sqrt{16*7}=\sqrt{16}*\sqrt{7}=4\sqrt{7}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-4\sqrt{7}}{2*6}=\frac{4-4\sqrt{7}}{12}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+4\sqrt{7}}{2*6}=\frac{4+4\sqrt{7}}{12}
$