What are the solutions to the equation x^2-4x+1=-5 ?

The solutions to the equation x^2-4x+1=-5 are x=2+sqrt(2)i,x=2-sqrt(2)i

Find the zeros of x^2-4x+1=-5

The zeros of x^2-4x+1=-5 are x=2+sqrt(2)i,x=2-sqrt(2)i

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x^{2} - 4 x + 1 = - 5

x = 2 + 2 i, x = 2 - 2 i

Solution steps

x^{2} - 4 x + 1 = - 5

Move

5

to the left side

x^{2} - 4 x + 6 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 4 ) \pm ( - 4 ) ^{2} - 4 \cdot 1 \cdot 6}{2 \cdot 1}

Simplify

(- 4)^{2} - 4 \cdot 1 \cdot 6 : 22 i

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

Separate the solutions

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

x = \frac{- ( - 4 ) + 2 2 i}{2 \cdot 1} : 2 + 2 i

x = \frac{- ( - 4 ) - 2 2 i}{2 \cdot 1} : 2 - 2 i

The solutions to the quadratic equation are:

x = 2 + 2 i, x = 2 - 2 i

(- \frac{1}{2}) x = 6

∣ t ∣ = 0

∣ 3 x + 2 ∣ = ∣ x - 5 ∣

\frac{5}{x - 2} < 0

- 3 \leq \frac{2 m + 6}{2} \leq 12

What are the solutions to the equation x^2-4x+1=-5 ?
The solutions to the equation x^2-4x+1=-5 are x=2+sqrt(2)i,x=2-sqrt(2)i
Find the zeros of x^2-4x+1=-5
The zeros of x^2-4x+1=-5 are x=2+sqrt(2)i,x=2-sqrt(2)i