What are the solutions to the equation 2x^0-13x^2+26x-24=0 ?

The solutions to the equation 2x^0-13x^2+26x-24=0 are x=1-i(3sqrt(13))/(13),x=1+i(3sqrt(13))/(13)

Find the zeros of 2x^0-13x^2+26x-24=0

The zeros of 2x^0-13x^2+26x-24=0 are x=1-i(3sqrt(13))/(13),x=1+i(3sqrt(13))/(13)

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2 x^{0} - 13 x^{2} + 26 x - 24 = 0

x = 1 - i \frac{3 13}{13}, x = 1 + i \frac{3 13}{13}

Solution steps

2 x^{0} - 13 x^{2} + 26 x - 24 = 0

Refine

- 13 x^{2} + 26 x - 22 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 26 \pm 2 6 ^{2} - 4 ( - 13 ) ( - 22 )}{2 ( - 13 )}

Simplify

2 6^{2} - 4 (- 13) (- 22) : 613 i

x_{1, 2} = \frac{- 26 \pm 6 13 i}{2 ( - 13 )}

Separate the solutions

x_{1} = \frac{- 26 + 6 13 i}{2 ( - 13 )}, x_{2} = \frac{- 26 - 6 13 i}{2 ( - 13 )}

x = \frac{- 26 + 6 13 i}{2 ( - 13 )} : 1 - i \frac{3 13}{13}

x = \frac{- 26 - 6 13 i}{2 ( - 13 )} : 1 + i \frac{3 13}{13}

The solutions to the quadratic equation are:

x = 1 - i \frac{3 13}{13}, x = 1 + i \frac{3 13}{13}

(d - x)^{2} = 2 x^{2}

2 x (4 x - 2) - 4 = 0

(a + b)^{2} x^{2} + (a - b)^{2} = 0

x, ma x z = 6 x^{1} + 4 x^{2}

x, f (x - 2) = x^{2} + 15

What are the solutions to the equation 2x^0-13x^2+26x-24=0 ?
The solutions to the equation 2x^0-13x^2+26x-24=0 are x=1-i(3sqrt(13))/(13),x=1+i(3sqrt(13))/(13)
Find the zeros of 2x^0-13x^2+26x-24=0
The zeros of 2x^0-13x^2+26x-24=0 are x=1-i(3sqrt(13))/(13),x=1+i(3sqrt(13))/(13)