Solve 4x^2 = 20x -24

Hello! I'm doing some review for a math placement test tomorrow. I can't remember how to solve this sort of equation:
Solve 4x^2 = 20x -24

I just know that it will have two answers. Please try to show the steps you used and I can probably figure out what you did.
Thanks for the help!

Bringing all the terms in lhs we get
4x^2 - 20x + 24 = 0
Divide the equation by 4 we get
x^2 - 5x +6 = 0
=> x^2 - 2x - 3x +6 = 0
=> x(x - 2) - 3(x - 2) = 0
=> (x - 2)(x - 3) = 0
=> x = 2, 3

this is a quadratic equation of the form
ax^2 +bx +c=0
bringing all the terms in lhs we get
4x^2 - 20x + 24 = 0
divide the equation by 4 we get
x^2 - 5x +6 = 0
i.e. x^2 - 2x - 3x +6 = 0
i.e. x(x - 2) - 3(x - 2) = 0
i.e. (x - 2)(x - 3) = 0
therefore x-2=0 and x-3=0
hence x=2 and x=3

4x² = 20x - 24
x² - 5/2x = - 6 + (- 5/2)²
x² - 5/2x = (- 24 + 25)/4
(x - 5/2)² = 1/4
x - 5/2 = ± 1/2

x = 5/2 - 1/2, x = 4/2, x = 2
x = 5/2 + 1/2, x = 6/2, x = 3

Answer: x = 2, 3

Proof of equality in two equations:
4(2)² = 20(2) - 24
4(4) = 40 - 24
16 = 16

4(3)² = 20(3) - 24
4(9) = 60 - 24
36 = 36

4x^2=20x-24

4x^2-20x+24=0

Then use the Quadratic equation:

x1,2 = -b ± sqrt( b^2-4ac )
------------------
2a

(a= 4; b= 20; and c=24

x=2
x=3

put in form so you can factor...
4x^2-20x+24=0 find factor of equations
(2x-4)(2x-6) take 2 out of whole eq..

2(x-2)(x-3) x=2or x=3

4x^2 - 20x + 24 = 0
x^2 - 5x + 6 = 0
x^2 - 3x - 2x + 6 = 0
x (x-3) -2 (x-3)

x=2 , x =3

4x^2 - 20x + 24

x^2 - 5x + 6

(x - 6)(x + 1)