What are the solutions to the equation p^2-12p+32=0 ?

The solutions to the equation p^2-12p+32=0 are p=8,p=4

Find the zeros of p^2-12p+32=0

The zeros of p^2-12p+32=0 are p=8,p=4

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p^{2} - 12 p + 32 = 0

p = 8, p = 4

Solution steps

p^{2} - 12 p + 32 = 0

Solve with the quadratic formula

p_{1, 2} = \frac{- ( - 12 ) \pm ( - 12 ) ^{2} - 4 \cdot 1 \cdot 32}{2 \cdot 1}

(- 12)^{2} - 4 \cdot 1 \cdot 32 = 4

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

Separate the solutions

p_{1} = \frac{- ( - 12 ) + 4}{2 \cdot 1}, p_{2} = \frac{- ( - 12 ) - 4}{2 \cdot 1}

p = \frac{- ( - 12 ) + 4}{2 \cdot 1} : 8

p = \frac{- ( - 12 ) - 4}{2 \cdot 1} : 4

The solutions to the quadratic equation are:

p = 8, p = 4

4 x^{2} + 25 = - 75

4 x^{2} - 15 = - 19

(x - \frac{( 2 5 )}{5})^{2} = - \frac{r}{30}

(x + 1) (x + 8) = 0

a (5 a - 3) = 2 (a^{2} + 5)

What are the solutions to the equation p^2-12p+32=0 ?
The solutions to the equation p^2-12p+32=0 are p=8,p=4
Find the zeros of p^2-12p+32=0
The zeros of p^2-12p+32=0 are p=8,p=4