What are the solutions to the equation x^2-10x+41=0 ?

The solutions to the equation x^2-10x+41=0 are x=5+4i,x=5-4i

Find the zeros of x^2-10x+41=0

The zeros of x^2-10x+41=0 are x=5+4i,x=5-4i

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x^{2} - 10 x + 41 = 0

x = 5 + 4 i, x = 5 - 4 i

Solution steps

x^{2} - 10 x + 41 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 10 ) \pm ( - 10 ) ^{2} - 4 \cdot 1 \cdot 41}{2 \cdot 1}

Simplify

(- 10)^{2} - 4 \cdot 1 \cdot 41 : 8 i

x_{1, 2} = \frac{- ( - 10 ) \pm 8 i}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 10 ) + 8 i}{2 \cdot 1}, x_{2} = \frac{- ( - 10 ) - 8 i}{2 \cdot 1}

x = \frac{- ( - 10 ) + 8 i}{2 \cdot 1} : 5 + 4 i

x = \frac{- ( - 10 ) - 8 i}{2 \cdot 1} : 5 - 4 i

The solutions to the quadratic equation are:

x = 5 + 4 i, x = 5 - 4 i

2 (y + 1) + 4 y = 3 (2 y - 1) + 8

- 38 d - 57 \geq - 19 d + 76

\frac{1}{( \frac{x}{y} )}

- \frac{3}{2} - 4

What are the solutions to the equation x^2-10x+41=0 ?
The solutions to the equation x^2-10x+41=0 are x=5+4i,x=5-4i
Find the zeros of x^2-10x+41=0
The zeros of x^2-10x+41=0 are x=5+4i,x=5-4i