What are the solutions to the equation a^2-18a+81=0 ?

The solutions to the equation a^2-18a+81=0 are a=9

Find the zeros of a^2-18a+81=0

The zeros of a^2-18a+81=0 are a=9

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

a = 9

Solution steps

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

Solve with the quadratic formula

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

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

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

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

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

a = 9

The solution to the quadratic equation is:

a = 9

(x - 7) (x - 3) = 0

x^{2} + 1.2 x = 1

x^{2} - 14 x = 11

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

x^{2} + 6 x - 1 = 0

What are the solutions to the equation a^2-18a+81=0 ?
The solutions to the equation a^2-18a+81=0 are a=9
Find the zeros of a^2-18a+81=0
The zeros of a^2-18a+81=0 are a=9