What are the solutions to the equation x^2-8x+12=0 ?

The solutions to the equation x^2-8x+12=0 are x=6,x=2

Find the zeros of x^2-8x+12=0

The zeros of x^2-8x+12=0 are x=6,x=2

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x^{2} - 8 x + 12 = 0

x = 6, x = 2

Solution steps

x^{2} - 8 x + 12 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 8 ) \pm ( - 8 ) ^{2} - 4 \cdot 1 \cdot 12}{2 \cdot 1}

(- 8)^{2} - 4 \cdot 1 \cdot 12 = 4

x_{1, 2} = \frac{- ( - 8 ) \pm 4}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 8 ) + 4}{2 \cdot 1}, x_{2} = \frac{- ( - 8 ) - 4}{2 \cdot 1}

x = \frac{- ( - 8 ) + 4}{2 \cdot 1} : 6

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

The solutions to the quadratic equation are:

x = 6, x = 2

28 > 4 m - 4

\frac{4 ^{7}}{4 ^{3}}

8 j^{3} + 4 j^{2} + 14 j + 7

4 (p - 4) - (p + 7) = 5 (p - 3)

(\frac{9}{5})^{- 1}

What are the solutions to the equation x^2-8x+12=0 ?
The solutions to the equation x^2-8x+12=0 are x=6,x=2
Find the zeros of x^2-8x+12=0
The zeros of x^2-8x+12=0 are x=6,x=2