What are the solutions to the equation x^2-18x+82=0 ?

The solutions to the equation x^2-18x+82=0 are x=9+i,x=9-i

Find the zeros of x^2-18x+82=0

The zeros of x^2-18x+82=0 are x=9+i,x=9-i

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x^{2} - 18 x + 82 = 0

x = 9 + i, x = 9 - i

Solution steps

x^{2} - 18 x + 82 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 18 ) \pm ( - 18 ) ^{2} - 4 \cdot 1 \cdot 82}{2 \cdot 1}

Simplify

(- 18)^{2} - 4 \cdot 1 \cdot 82 : 2 i

x_{1, 2} = \frac{- ( - 18 ) \pm 2 i}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 18 ) + 2 i}{2 \cdot 1}, x_{2} = \frac{- ( - 18 ) - 2 i}{2 \cdot 1}

x = \frac{- ( - 18 ) + 2 i}{2 \cdot 1} : 9 + i

x = \frac{- ( - 18 ) - 2 i}{2 \cdot 1} : 9 - i

The solutions to the quadratic equation are:

x = 9 + i, x = 9 - i

x^{2} = x + 20

- 16 t^{2} + 96 t + 112

3 x^{2} + 6 x - 62 = 7

t^{2} + t + 1 = (t - 1)^{2}

2 b^{3} + b^{5}

What are the solutions to the equation x^2-18x+82=0 ?
The solutions to the equation x^2-18x+82=0 are x=9+i,x=9-i
Find the zeros of x^2-18x+82=0
The zeros of x^2-18x+82=0 are x=9+i,x=9-i