What are the solutions to the equation x(x-12)+32=0 ?

The solutions to the equation x(x-12)+32=0 are x=8,x=4

Find the zeros of x(x-12)+32=0

The zeros of x(x-12)+32=0 are x=8,x=4

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x (x - 12) + 32 = 0

x = 8, x = 4

Solution steps

x (x - 12) + 32 = 0

Expand

x (x - 12) + 32 : x^{2} - 12 x + 32

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

Solve with the quadratic formula

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

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

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

x = 8, x = 4

x, x^{2} - x y + 5 y = 0

x, s (- 1) = x^{2} - 4 x - 21

x^{2} + 10 x + 37 = 0

(x + 7) (x + 2) = 0

x^{2} = 20 x - 51

What are the solutions to the equation x(x-12)+32=0 ?
The solutions to the equation x(x-12)+32=0 are x=8,x=4
Find the zeros of x(x-12)+32=0
The zeros of x(x-12)+32=0 are x=8,x=4