What are the solutions to the equation x^2-18x+81=16 ?

The solutions to the equation x^2-18x+81=16 are x=13,x=5

Find the zeros of x^2-18x+81=16

The zeros of x^2-18x+81=16 are x=13,x=5

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x^{2} - 18 x + 81 = 16

x = 13, x = 5

Solution steps

x^{2} - 18 x + 81 = 16

Move

16

to the left side

x^{2} - 18 x + 65 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 18 ) \pm ( - 18 ) ^{2} - 4 \cdot 1 \cdot 65}{2 \cdot 1}

(- 18)^{2} - 4 \cdot 1 \cdot 65 = 8

x_{1, 2} = \frac{- ( - 18 ) \pm 8}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 18 ) + 8}{2 \cdot 1}, x_{2} = \frac{- ( - 18 ) - 8}{2 \cdot 1}

x = \frac{- ( - 18 ) + 8}{2 \cdot 1} : 13

x = \frac{- ( - 18 ) - 8}{2 \cdot 1} : 5

The solutions to the quadratic equation are:

x = 13, x = 5

9 r (r - 2) = r^{2} + 5

x^{2} + 8 x + 3 = 0

16 x^{2} + 24 x + 5 = 0

y, - 3 y^{2} z - 3 y z^{2} = 21

p^{2} + 14 p + 49 = 2

What are the solutions to the equation x^2-18x+81=16 ?
The solutions to the equation x^2-18x+81=16 are x=13,x=5
Find the zeros of x^2-18x+81=16
The zeros of x^2-18x+81=16 are x=13,x=5