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

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

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

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

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

p = 6

Solution steps

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

Solve with the quadratic formula

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

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

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

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

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

p = 6

The solution to the quadratic equation is:

p = 6

X^{2} - X - 4 = 0

x^{2} - 7 x - 30 = 0

\frac{1}{2} x^{2} - 2 x = 6

x, z = 15 x^{2} + 23 x y - 14 y^{2}

3 x^{2} - 5 = - 8 x - 10

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