What are the solutions to the equation z^2-12z+36=0 ?

The solutions to the equation z^2-12z+36=0 are z=6

Find the zeros of z^2-12z+36=0

The zeros of z^2-12z+36=0 are z=6

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z^{2} - 12 z + 36 = 0

z = 6

Solution steps

z^{2} - 12 z + 36 = 0

Solve with the quadratic formula

z_{1, 2} = \frac{- ( - 12 ) \pm ( - 12 ) ^{2} - 4 \cdot 1 \cdot 36}{2 \cdot 1}

(- 12)^{2} - 4 \cdot 1 \cdot 36 = 0

z_{1, 2} = \frac{- ( - 12 ) \pm 0}{2 \cdot 1}

z = \frac{- ( - 12 )}{2 \cdot 1}

\frac{- ( - 12 )}{2 \cdot 1} = 6

z = 6

The solution to the quadratic equation is:

z = 6

x^{2} - x - 56 = 0

x^{2} + 4 x - 21 = 0

0 = x^{2} + 8

2 x^{2} - 8 x + 12 = - 3 x

8 x^{2} + 7 x = - 2

What are the solutions to the equation z^2-12z+36=0 ?
The solutions to the equation z^2-12z+36=0 are z=6
Find the zeros of z^2-12z+36=0
The zeros of z^2-12z+36=0 are z=6