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

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

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

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

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

x = 8, x = 4

Solution steps

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

Switch sides

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

0 = 2 x^{2} - x - 10

x^{2} + 2.2 x - 10.35 = 0

x, f = 5 x^{2} + 6 y^{3} + 2 z^{4}

z^{2} - 14 z = - 17

3 x^{2} - 6 x - 15 = 0

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