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solve for x: (4x-3)^2=-15

Mathematics
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so (4x-3)^2=(4x-3)(4x-3) right
yes
so take that and foil it

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Other answers:

(4x-3)(4x-3) = 4x(4x - 3) - 3(4x - 3) = 16x^2 - 12x - 12x +9 = 16x^2 - 24x + 9
16x^2-24x+24=0
but its not = to 0 its =-15
(4x-3)(4x-3) = 4x(4x - 3) - 3(4x - 3) = 16x^2 - 12x - 12x +9 = 16x^2 - 24x + 9
my bad see you already did that
so now is there a common number that they are all divisible by?
wait im at 16x^2-24x+9=-15 right?
ya then add the 15 to get 16x^2-24x+24=0
okay yeah i did that
now find a common denominator
8?
yup
so does that then turn into 2x^2-3x=3=0?
2x^2-3x+3 sorry
16x^2-24x+9=-15 16x^2 - 24x + 24 = 0 8(2x^2 - 3x + 3) = 0 2x^2 - 3x + 3 = 0 x^2 - 3/2 + 3/2 = 0 x^2 - 3/2x = -3/2 x^2 -3/2x + 9/16 = -3/2 + 9/16 (x - 3/4)^2 = -15/16 Solution is imaginary
yes
then what do i do?
the answer is 3 + or - the square root of 15 divided by4 but idk how to get it
i would do the quadratic formula
-b + or - the square root of b^2 -4ac divided by 2a?
\[\frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]
(-b+-sqrt(b^2-4ac))/2a
yup
i got it thanks to both of you!
not a problem
You got what? Mind sharing what you got?
3 + or - the square root of 15 divided by 4
3+ or - i square root 15 divided by 4 sorry i forgot the "i"
yup

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