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

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

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

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

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

x = 6

Solution steps

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

Solve with the quadratic formula

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

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

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

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

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

x = 6

The solution to the quadratic equation is:

x = 6

\frac{2 - 5}{2}

∣ - 3 x + 2 ∣ > 4

3 x + 4 \geq 16

- 12 = 10 n

x + 2 (8 - x) = 15

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