How do you solve 12x^2-5=17x through factoring?

Write as 12x^2-17x-5=0

Factor 12x^2-17x-5 - we have to find factors of 12 for the first terms and factors of 5 for the second.

You need to go through all the possibilities in order to make the first term multiply to 12x^2 and the constant term multiply to 5.

(x-5)(12x+1)
(x-1)(12x+5)
(x+5)(12x-1)
(x+1)(12x-5)
(2x-5)(6x+1)
(2x-1)(6x+5)

eventually you start to see when certain groupings won't get us to -17x in the middle. The winner is

(4x+1)(3x-5)

Admittedly, this solution did not just jump out at me either. Sometimes these factoring problems are tedious.

12x^2 - 5 = 17x
12x^2 - 17x - 5 = 0
12x^2 + 3x - 20x - 5 = 0
(12x^2 + 3x) - (20x + 5) = 0
3x(4x + 1) - 5(4x + 1) = 0
(4x + 1)(3x - 5) = 0

4x + 1 = 0
4x = -1
x = -1/4 (-0.25)

3x - 5 = 0
3x = 5
x = 5/3

∴ x = -1/4 (-0.25), 5/3

12x^2 - 17x - 5 = 0

12x^2-20x+3x-5=0

4x(3x-5)+1(3x-5)

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

x = - 1/4 , x = 5/3

12x^2 - 17x - 5 = 0

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

x = - 1/4 , x = 5/3

12x^2 - 5 = 17x

12x^2 - 17x - 5 = 0

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

4x + 1 = 0 implies 4x = -1, or x = -1/4

3x - 5 = 0 imples 3x = 5, or x = 5/3

x = -1/4 and x = 5/3 are the answers

12x^2 - 17x - 5 = 0
(4x + 1) (3x - 5) = 0

x = -1/4 or x = +5/3

12x² - 5 = 17x
12x² - 17x - 5 = 0
(3x - 5)(4x + 1) = 0

x = 5/3; x = -1/4