What are the solutions to the equation 8z^2-9z+16=0 ?

The solutions to the equation 8z^2-9z+16=0 are z= 9/16+i(sqrt(431))/(16),z= 9/16-i(sqrt(431))/(16)

Find the zeros of 8z^2-9z+16=0

The zeros of 8z^2-9z+16=0 are z= 9/16+i(sqrt(431))/(16),z= 9/16-i(sqrt(431))/(16)

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8 z^{2} - 9 z + 16 = 0

z = \frac{9}{16} + i \frac{431}{16}, z = \frac{9}{16} - i \frac{431}{16}

Solution steps

8 z^{2} - 9 z + 16 = 0

Solve with the quadratic formula

z_{1, 2} = \frac{- ( - 9 ) \pm ( - 9 ) ^{2} - 4 \cdot 8 \cdot 16}{2 \cdot 8}

Simplify

(- 9)^{2} - 4 \cdot 8 \cdot 16 : 431 i

z_{1, 2} = \frac{- ( - 9 ) \pm 431 i}{2 \cdot 8}

Separate the solutions

z_{1} = \frac{- ( - 9 ) + 431 i}{2 \cdot 8}, z_{2} = \frac{- ( - 9 ) - 431 i}{2 \cdot 8}

z = \frac{- ( - 9 ) + 431 i}{2 \cdot 8} : \frac{9}{16} + i \frac{431}{16}

z = \frac{- ( - 9 ) - 431 i}{2 \cdot 8} : \frac{9}{16} - i \frac{431}{16}

The solutions to the quadratic equation are:

z = \frac{9}{16} + i \frac{431}{16}, z = \frac{9}{16} - i \frac{431}{16}

- 6 m^{2} = 48

(x - 1)^{2} = - 7

x^{2} + 2 x + 1 = 0

(x - 1)^{2} = 13

v^{2} - 8 v + 7 = 0

What are the solutions to the equation 8z^2-9z+16=0 ?
The solutions to the equation 8z^2-9z+16=0 are z= 9/16+i(sqrt(431))/(16),z= 9/16-i(sqrt(431))/(16)
Find the zeros of 8z^2-9z+16=0
The zeros of 8z^2-9z+16=0 are z= 9/16+i(sqrt(431))/(16),z= 9/16-i(sqrt(431))/(16)