What are the solutions to the equation 16z^2-64z+73=0 ?

The solutions to the equation 16z^2-64z+73=0 are z=2+i 3/4 ,z=2-i 3/4

Find the zeros of 16z^2-64z+73=0

The zeros of 16z^2-64z+73=0 are z=2+i 3/4 ,z=2-i 3/4

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16 z^{2} - 64 z + 73 = 0

z = 2 + i \frac{3}{4}, z = 2 - i \frac{3}{4}

Solution steps

16 z^{2} - 64 z + 73 = 0

Solve with the quadratic formula

z_{1, 2} = \frac{- ( - 64 ) \pm ( - 64 ) ^{2} - 4 \cdot 16 \cdot 73}{2 \cdot 16}

Simplify

(- 64)^{2} - 4 \cdot 16 \cdot 73 : 24 i

z_{1, 2} = \frac{- ( - 64 ) \pm 24 i}{2 \cdot 16}

Separate the solutions

z_{1} = \frac{- ( - 64 ) + 24 i}{2 \cdot 16}, z_{2} = \frac{- ( - 64 ) - 24 i}{2 \cdot 16}

z = \frac{- ( - 64 ) + 24 i}{2 \cdot 16} : 2 + i \frac{3}{4}

z = \frac{- ( - 64 ) - 24 i}{2 \cdot 16} : 2 - i \frac{3}{4}

The solutions to the quadratic equation are:

z = 2 + i \frac{3}{4}, z = 2 - i \frac{3}{4}

2 x^{2} + 2 + 5 x = - 4 - x

2 b + 4 = - 18 - 9 b

1 - a + 6 a > 6

\frac{4}{6.02 \cdot 1 0 ^{23}}

- 8 \cdot (- 10) - 7 \cdot \frac{1}{- 1}

What are the solutions to the equation 16z^2-64z+73=0 ?
The solutions to the equation 16z^2-64z+73=0 are z=2+i 3/4 ,z=2-i 3/4
Find the zeros of 16z^2-64z+73=0
The zeros of 16z^2-64z+73=0 are z=2+i 3/4 ,z=2-i 3/4