solve for x

(19x-34)^2 + 68 = 68x

im trying to solve this in long method but i seems like the method is hard to factor

(19x)^2 - 2(19x)(34)x + 34^2 + 68 = 38x

is this any short method to find the x?

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- Dec 22nd 2012, 07:16 AMrcsvalue of x
solve for x

(19x-34)^2 + 68 = 68x

im trying to solve this in long method but i seems like the method is hard to factor

(19x)^2 - 2(19x)(34)x + 34^2 + 68 = 38x

is this any short method to find the x? - Dec 22nd 2012, 07:21 AMPlatoRe: value of x
- Dec 22nd 2012, 07:29 AMHallsofIvyRe: value of x
What you have so far is incorrect but probably just a typo- you should have 2(19)(34)x. There is no "x" with the 19. In any case, I can see no reason not to finish it:

361x^2- 1292x+ 1156= 38x which gives 361x^2- 1330x+ 1156= 0.

If you cannot factor that, use the quadratic formula. - Dec 22nd 2012, 07:39 AMrcsRe: value of x
thanks Guys