What are the solutions to the equation v^2-4v+26=4 ?

The solutions to the equation v^2-4v+26=4 are v=2+3sqrt(2)i,v=2-3sqrt(2)i

Find the zeros of v^2-4v+26=4

The zeros of v^2-4v+26=4 are v=2+3sqrt(2)i,v=2-3sqrt(2)i

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v^{2} - 4 v + 26 = 4

v = 2 + 32 i, v = 2 - 32 i

Solution steps

v^{2} - 4 v + 26 = 4

Move

4

to the left side

v^{2} - 4 v + 22 = 0

Solve with the quadratic formula

v_{1, 2} = \frac{- ( - 4 ) \pm ( - 4 ) ^{2} - 4 \cdot 1 \cdot 22}{2 \cdot 1}

Simplify

(- 4)^{2} - 4 \cdot 1 \cdot 22 : 62 i

v_{1, 2} = \frac{- ( - 4 ) \pm 6 2 i}{2 \cdot 1}

Separate the solutions

v_{1} = \frac{- ( - 4 ) + 6 2 i}{2 \cdot 1}, v_{2} = \frac{- ( - 4 ) - 6 2 i}{2 \cdot 1}

v = \frac{- ( - 4 ) + 6 2 i}{2 \cdot 1} : 2 + 32 i

v = \frac{- ( - 4 ) - 6 2 i}{2 \cdot 1} : 2 - 32 i

The solutions to the quadratic equation are:

v = 2 + 32 i, v = 2 - 32 i

12 x = x^{2}

- 6 x^{2} + 7 x + 3 = 0

(k - 2) x^{2} - (2 k - 1) x + (k - 1) = 0

3 x^{2} - 12 x + 24 = 0

x^{2} - 6 x - 1

What are the solutions to the equation v^2-4v+26=4 ?
The solutions to the equation v^2-4v+26=4 are v=2+3sqrt(2)i,v=2-3sqrt(2)i
Find the zeros of v^2-4v+26=4
The zeros of v^2-4v+26=4 are v=2+3sqrt(2)i,v=2-3sqrt(2)i