33(33)=(2x-12)(2x+10)

Solution for 33(33)=(2x-12)(2x+10) equation:

33(33)=(2x-12)(2x+10)

We move all terms to the left:

33(33)-((2x-12)(2x+10))=0

We multiply parentheses ..

-((+4x^2+20x-24x-120))+3333=0


We calculate terms in parentheses: -((+4x^2+20x-24x-120)), so:

(+4x^2+20x-24x-120)

We get rid of parentheses

4x^2+20x-24x-120

We add all the numbers together, and all the variables

4x^2-4x-120

Back to the equation:

-(4x^2-4x-120)


We get rid of parentheses

-4x^2+4x+120+3333=0

We add all the numbers together, and all the variables

-4x^2+4x+3453=0

a = -4; b = 4; c = +3453;
Δ = b2-4ac
Δ = 42-4·(-4)·3453
Δ = 55264
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:

$\sqrt{\Delta}=\sqrt{55264}=\sqrt{16*3454}=\sqrt{16}*\sqrt{3454}=4\sqrt{3454}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-4\sqrt{3454}}{2*-4}=\frac{-4-4\sqrt{3454}}{-8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+4\sqrt{3454}}{2*-4}=\frac{-4+4\sqrt{3454}}{-8}
$