What are the solutions to the equation 2z^2-16z=-160 ?

The solutions to the equation 2z^2-16z=-160 are z=4+8i,z=4-8i

Find the zeros of 2z^2-16z=-160

The zeros of 2z^2-16z=-160 are z=4+8i,z=4-8i

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2 z^{2} - 16 z = - 160

z = 4 + 8 i, z = 4 - 8 i

Solution steps

2 z^{2} - 16 z = - 160

Move

160

to the left side

2 z^{2} - 16 z + 160 = 0

Solve with the quadratic formula

z_{1, 2} = \frac{- ( - 16 ) \pm ( - 16 ) ^{2} - 4 \cdot 2 \cdot 160}{2 \cdot 2}

Simplify

(- 16)^{2} - 4 \cdot 2 \cdot 160 : 32 i

z_{1, 2} = \frac{- ( - 16 ) \pm 32 i}{2 \cdot 2}

Separate the solutions

z_{1} = \frac{- ( - 16 ) + 32 i}{2 \cdot 2}, z_{2} = \frac{- ( - 16 ) - 32 i}{2 \cdot 2}

z = \frac{- ( - 16 ) + 32 i}{2 \cdot 2} : 4 + 8 i

z = \frac{- ( - 16 ) - 32 i}{2 \cdot 2} : 4 - 8 i

The solutions to the quadratic equation are:

z = 4 + 8 i, z = 4 - 8 i

2 x^{2} + 6 x - 1

x^{2} + 2 x - 44 = 0

- 2 x^{2} = - 42

7 x^{2} + 14 x - 56 = 0

x^{2} = \frac{64}{81}

What are the solutions to the equation 2z^2-16z=-160 ?
The solutions to the equation 2z^2-16z=-160 are z=4+8i,z=4-8i
Find the zeros of 2z^2-16z=-160
The zeros of 2z^2-16z=-160 are z=4+8i,z=4-8i