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

The solutions to the equation x^2+81=18x are x=9

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

The zeros of x^2+81=18x are x=9

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x^{2} + 81 = 18 x

x = 9

Solution steps

x^{2} + 81 = 18 x

Move

18 x

to the left side

x^{2} + 81 - 18 x = 0

Write in the standard form

a x^{2} + b x + c = 0

x^{2} - 18 x + 81 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 18 ) \pm ( - 18 ) ^{2} - 4 \cdot 1 \cdot 81}{2 \cdot 1}

(- 18)^{2} - 4 \cdot 1 \cdot 81 = 0

x_{1, 2} = \frac{- ( - 18 ) \pm 0}{2 \cdot 1}

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

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

x = 9

The solution to the quadratic equation is:

x = 9

6 x^{2} - 4 = 2 x^{2} + 8

x, 4 x^{2} - 12 x y + 9 y^{2} = 0

k^{2} + 9 k + 14 = 0

- 32 x^{2} + 98 = 0

3 z^{2} + 6 z + 3 = 0

What are the solutions to the equation x^2+81=18x ?
The solutions to the equation x^2+81=18x are x=9
Find the zeros of x^2+81=18x
The zeros of x^2+81=18x are x=9