What are the solutions to the equation x^2-8x+32=0 ?

The solutions to the equation x^2-8x+32=0 are x=4+4i,x=4-4i

Find the zeros of x^2-8x+32=0

The zeros of x^2-8x+32=0 are x=4+4i,x=4-4i

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x^{2} - 8 x + 32 = 0

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

Solution steps

x^{2} - 8 x + 32 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 8 ) \pm ( - 8 ) ^{2} - 4 \cdot 1 \cdot 32}{2 \cdot 1}

Simplify

(- 8)^{2} - 4 \cdot 1 \cdot 32 : 8 i

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

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

100 \cdot \frac{π}{180}

5 x - (x + 2) \geq 3 (x - 1) + 6

2 (x + 4) - x (x - 5) = 7 (x - 8)

- 4 x + 2 \geq 26

(0)^{2} = v + 9.8 (0.9)

What are the solutions to the equation x^2-8x+32=0 ?
The solutions to the equation x^2-8x+32=0 are x=4+4i,x=4-4i
Find the zeros of x^2-8x+32=0
The zeros of x^2-8x+32=0 are x=4+4i,x=4-4i