How do you solve $4/3x^2 - 2x + 3/4 = 0$ using the quadratic formula?

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Answer 1 · 2018-06-02T22:07:52

x=3/4

$ax^2+bx+c$

$4/3x^2 - 2x + 3/4 = 0$

I would start by multiplying the whole thing by the common denominator of the fractions: $12$ then we can just deal with integers.

$12(4/3x^2 - 2x + 3/4 = 0)$

$16x^2 -24x +9 =0$

$a=16, b=-24, c=9$

$x=(-b+-sqrt(b^2-4ac))/(2a)$

$x=(-(-24)+-sqrt((-24)^2-4*16*9))/(2*16)$

$x=(24+-sqrt(576-576))/(32)$

$x=24/32=3/4$

graph{16x^2 -24x +9 [-9.54, 10.46, -1.2, 8.8]}

How do you solve 4/3x^2 - 2x + 3/4 = 0 4/3x^2 - 2x + 3/4 = 0 using the quadratic formula?