What are the solutions to the equation x^2-4x+29=0 ?

The solutions to the equation x^2-4x+29=0 are x=2+5i,x=2-5i

Find the zeros of x^2-4x+29=0

The zeros of x^2-4x+29=0 are x=2+5i,x=2-5i

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x^{2} - 4 x + 29 = 0

x = 2 + 5 i, x = 2 - 5 i

Solution steps

x^{2} - 4 x + 29 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 4 ) \pm ( - 4 ) ^{2} - 4 \cdot 1 \cdot 29}{2 \cdot 1}

Simplify

(- 4)^{2} - 4 \cdot 1 \cdot 29 : 10 i

x_{1, 2} = \frac{- ( - 4 ) \pm 10 i}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 4 ) + 10 i}{2 \cdot 1}, x_{2} = \frac{- ( - 4 ) - 10 i}{2 \cdot 1}

x = \frac{- ( - 4 ) + 10 i}{2 \cdot 1} : 2 + 5 i

x = \frac{- ( - 4 ) - 10 i}{2 \cdot 1} : 2 - 5 i

The solutions to the quadratic equation are:

x = 2 + 5 i, x = 2 - 5 i

4 n + 2 = 6 (\frac{1}{3} n - \frac{2}{3})

\frac{- 20 r ^{5} s ^{4}}{16 r s ^{8}}

1 < x \leq 3

3 x^{2} = 16 - 8 x

lo g_{3} (0.2)

What are the solutions to the equation x^2-4x+29=0 ?
The solutions to the equation x^2-4x+29=0 are x=2+5i,x=2-5i
Find the zeros of x^2-4x+29=0
The zeros of x^2-4x+29=0 are x=2+5i,x=2-5i