What are the solutions to the equation z^2-14z+49=-121 ?

The solutions to the equation z^2-14z+49=-121 are z=7+11i,z=7-11i

Find the zeros of z^2-14z+49=-121

The zeros of z^2-14z+49=-121 are z=7+11i,z=7-11i

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z^{2} - 14 z + 49 = - 121

z = 7 + 11 i, z = 7 - 11 i

Solution steps

z^{2} - 14 z + 49 = - 121

Move

121

to the left side

z^{2} - 14 z + 170 = 0

Solve with the quadratic formula

z_{1, 2} = \frac{- ( - 14 ) \pm ( - 14 ) ^{2} - 4 \cdot 1 \cdot 170}{2 \cdot 1}

Simplify

(- 14)^{2} - 4 \cdot 1 \cdot 170 : 22 i

z_{1, 2} = \frac{- ( - 14 ) \pm 22 i}{2 \cdot 1}

Separate the solutions

z_{1} = \frac{- ( - 14 ) + 22 i}{2 \cdot 1}, z_{2} = \frac{- ( - 14 ) - 22 i}{2 \cdot 1}

z = \frac{- ( - 14 ) + 22 i}{2 \cdot 1} : 7 + 11 i

z = \frac{- ( - 14 ) - 22 i}{2 \cdot 1} : 7 - 11 i

The solutions to the quadratic equation are:

z = 7 + 11 i, z = 7 - 11 i

3 n^{2} + 5 = - 46

x^{2} + 15 x - 23 = 0

5 t^{2} + 60 = 0

x^{2} + x + 2.5 = 0

a x^{2} - x = 0

What are the solutions to the equation z^2-14z+49=-121 ?
The solutions to the equation z^2-14z+49=-121 are z=7+11i,z=7-11i
Find the zeros of z^2-14z+49=-121
The zeros of z^2-14z+49=-121 are z=7+11i,z=7-11i