how do you factor 10x^2 - 19x + 6 = 0?

Answer 1

(5x-2)(2x-3)

So, x= 2/5 or 3/2

Answer 2

Use cross multiplication
factors of 10 are 1 and 10, or 2 and 5, -1- and 10, -2 and -5)
factors of 6 can be 1 and 6, or 2 and 3 ,-1 and -6, -2 and -3.
Use cross multiplication is such a way that the sum of the cross multiplied terms is -19
try ...............5x... -2 ......5 * the -3 below = -15
and...............2x... -3 .....2* the -2 above = -4
factors of 10...^.....^ factors of 6.......sum= -19 (coefficent of x)
so u got the correct factors.
Now, since 10 is the coefficient of x2, its factors have to have an x
so it is 5x and 2x in the cross mutiplication above
since using cross multiplication you have found the factors, you take the expressions on the same line 5x-2 and 2x-3
So these are your factors (5x-2)(2x-3)=0
Do not expect to choose the right factors to test out on the first try. Choose a combination and try, of course taking the combination that looks most likely to get the coefficient of x.

Hope this answer helped and is easy enough to understand,

Answer 3

(5x-2)(2x-3) = 0
multiply that out and you get 10x^2 - 19x + 6 = 0

Answer 4

10x^2-19x+6=0
=>10x^2-15x-4x+6=0
=>5x(2x-3)-2(2x-3)=0
=>(2x-3)(5x-2)=0
If you want to solve for x,
Either,
2x-3=0
=>2x=3
therefore, x=3/2

Or,
5x-2=0
=>5x=2
therefore, x=2/5

Answer 5

10x^-19x+6 = (5x-2)(2x-3)

Answer 6

10x^2-19x+6=0
10x^2-15x-4x+6=0
=>5x(2x-3)-2(2x-3)=0
=>(2x-3)(5x-2)=0
Hence either 2x-3=0 or 5x-2=0
x=3/2 or 1.5 and 2/5 or 0.4

Answer 7

Use Quadratic Equation

x= - (-19)+-sqrt((-19)^2-4(10)(6))

Answer 8

Do you this theorem: x=[ -b +/- (Sqrt (b^2-4ac))]/2a

Well use it, being a=10, b=-19 and c =6

so, x=[ 19+/- (Sqrt (361-240))] / 20

x= [19 +/- (Sqrt 121)]/20
x = [19 +11]/20 or x= [19-11]/20
x= 30/20 or x= 8/20
x= 3/2 or x= 2/5

Factorizing

(x-3/2) (x-2/5) =0

Answer 9

(5x-2)(2x-3)
x=2/5 x=3/2

Answer 10

(5x-2)(2x-3)

So x=2/5 or 3/2

Right?