5x(2x-4)=12-(3-2x)

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

5x(2x-4)=12-(3-2x)

We move all terms to the left:

5x(2x-4)-(12-(3-2x))=0

We add all the numbers together, and all the variables

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

We multiply parentheses

10x^2-20x-(12-(-2x+3))=0


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

12-(-2x+3)

determiningTheFunctionDomain
-(-2x+3)+12

We get rid of parentheses

2x-3+12

We add all the numbers together, and all the variables

2x+9

Back to the equation:

-(2x+9)


We get rid of parentheses

10x^2-20x-2x-9=0

We add all the numbers together, and all the variables

10x^2-22x-9=0

a = 10; b = -22; c = -9;
Δ = b2-4ac
Δ = -222-4·10·(-9)
Δ = 844
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:

$\sqrt{\Delta}=\sqrt{844}=\sqrt{4*211}=\sqrt{4}*\sqrt{211}=2\sqrt{211}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-22)-2\sqrt{211}}{2*10}=\frac{22-2\sqrt{211}}{20}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-22)+2\sqrt{211}}{2*10}=\frac{22+2\sqrt{211}}{20}
$