(3x-2)2-(2x+4)(2x-4)-5(x+1)2=-11(2x-1)

Solution for (3x-2)2-(2x+4)(2x-4)-5(x+1)2=-11(2x-1) equation:

(3x-2)2-(2x+4)(2x-4)-5(x+1)2=-11(2x-1)

We move all terms to the left:

(3x-2)2-(2x+4)(2x-4)-5(x+1)2-(-11(2x-1))=0

We use the square of the difference formula

4x^2+(3x-2)2-5(x+1)2-(-11(2x-1))+16=0

We multiply parentheses

4x^2+6x-10x-(-11(2x-1))-4-10+16=0


We calculate terms in parentheses: -(-11(2x-1)), so:

-11(2x-1)

We multiply parentheses

-22x+11

Back to the equation:

-(-22x+11)


We add all the numbers together, and all the variables

4x^2-4x-(-22x+11)+2=0

We get rid of parentheses

4x^2-4x+22x-11+2=0

We add all the numbers together, and all the variables

4x^2+18x-9=0

a = 4; b = 18; c = -9;
Δ = b2-4ac
Δ = 182-4·4·(-9)
Δ = 468
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:

$\sqrt{\Delta}=\sqrt{468}=\sqrt{36*13}=\sqrt{36}*\sqrt{13}=6\sqrt{13}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(18)-6\sqrt{13}}{2*4}=\frac{-18-6\sqrt{13}}{8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(18)+6\sqrt{13}}{2*4}=\frac{-18+6\sqrt{13}}{8}
$