What are the solutions to the equation z^2-3/4 z=-80 ?

The solutions to the equation z^2-3/4 z=-80 are z= 3/8+i(sqrt(5111))/8 ,z= 3/8-i(sqrt(5111))/8

Find the zeros of z^2-3/4 z=-80

The zeros of z^2-3/4 z=-80 are z= 3/8+i(sqrt(5111))/8 ,z= 3/8-i(sqrt(5111))/8

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z^{2} - \frac{3}{4} z = - 80

z = \frac{3}{8} + i \frac{5111}{8}, z = \frac{3}{8} - i \frac{5111}{8}

Solution steps

z^{2} - \frac{3}{4} z = - 80

Multiply both sides by

4

4 z^{2} - 3 z = - 320

Move

320

to the left side

4 z^{2} - 3 z + 320 = 0

Solve with the quadratic formula

z_{1, 2} = \frac{- ( - 3 ) \pm ( - 3 ) ^{2} - 4 \cdot 4 \cdot 320}{2 \cdot 4}

Simplify

(- 3)^{2} - 4 \cdot 4 \cdot 320 : 5111 i

z_{1, 2} = \frac{- ( - 3 ) \pm 5111 i}{2 \cdot 4}

Separate the solutions

z_{1} = \frac{- ( - 3 ) + 5111 i}{2 \cdot 4}, z_{2} = \frac{- ( - 3 ) - 5111 i}{2 \cdot 4}

z = \frac{- ( - 3 ) + 5111 i}{2 \cdot 4} : \frac{3}{8} + i \frac{5111}{8}

z = \frac{- ( - 3 ) - 5111 i}{2 \cdot 4} : \frac{3}{8} - i \frac{5111}{8}

The solutions to the quadratic equation are:

z = \frac{3}{8} + i \frac{5111}{8}, z = \frac{3}{8} - i \frac{5111}{8}

x^{1} + 5 x^{2} = 7

25 y^{2} - 25 y + 6 = 0

2 x^{2} - x = \frac{1}{2}

5 q^{2} = 18 q + 8

x^{2} + 7 x + 19 = 0

What are the solutions to the equation z^2-3/4 z=-80 ?
The solutions to the equation z^2-3/4 z=-80 are z= 3/8+i(sqrt(5111))/8 ,z= 3/8-i(sqrt(5111))/8
Find the zeros of z^2-3/4 z=-80
The zeros of z^2-3/4 z=-80 are z= 3/8+i(sqrt(5111))/8 ,z= 3/8-i(sqrt(5111))/8