What are the solutions to the equation z^2-4z-5=4z ?

The solutions to the equation z^2-4z-5=4z are z=4+sqrt(21),z=4-sqrt(21)

Find the zeros of z^2-4z-5=4z

The zeros of z^2-4z-5=4z are z=4+sqrt(21),z=4-sqrt(21)

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z^{2} - 4 z - 5 = 4 z

z = 4 + 21, z = 4 - 21

Decimal Notation

z = 8.58257 \dots, z = - 0.58257 \dots

Solution steps

z^{2} - 4 z - 5 = 4 z

Move

4 z

to the left side

z^{2} - 8 z - 5 = 0

Solve with the quadratic formula

z_{1, 2} = \frac{- ( - 8 ) \pm ( - 8 ) ^{2} - 4 \cdot 1 \cdot ( - 5 )}{2 \cdot 1}

(- 8)^{2} - 4 \cdot 1 \cdot (- 5) = 221

z_{1, 2} = \frac{- ( - 8 ) \pm 2 21}{2 \cdot 1}

Separate the solutions

z_{1} = \frac{- ( - 8 ) + 2 21}{2 \cdot 1}, z_{2} = \frac{- ( - 8 ) - 2 21}{2 \cdot 1}

z = \frac{- ( - 8 ) + 2 21}{2 \cdot 1} : 4 + 21

z = \frac{- ( - 8 ) - 2 21}{2 \cdot 1} : 4 - 21

The solutions to the quadratic equation are:

z = 4 + 21, z = 4 - 21

2 x (x - 2) = 0

8 t = - (16 t^{2} + 1)

x^{2} + x - 18 = 0

50 x^{2} - 200 x - 250 = 0

x^{2} + (\frac{3}{4})^{2} = 1

What are the solutions to the equation z^2-4z-5=4z ?
The solutions to the equation z^2-4z-5=4z are z=4+sqrt(21),z=4-sqrt(21)
Find the zeros of z^2-4z-5=4z
The zeros of z^2-4z-5=4z are z=4+sqrt(21),z=4-sqrt(21)