What are the solutions to the equation x^2-16x+28=0 ?

The solutions to the equation x^2-16x+28=0 are x=14,x=2

Find the zeros of x^2-16x+28=0

The zeros of x^2-16x+28=0 are x=14,x=2

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x^{2} - 16 x + 28 = 0

x = 14, x = 2

Solution steps

x^{2} - 16 x + 28 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 16 ) \pm ( - 16 ) ^{2} - 4 \cdot 1 \cdot 28}{2 \cdot 1}

(- 16)^{2} - 4 \cdot 1 \cdot 28 = 12

x_{1, 2} = \frac{- ( - 16 ) \pm 12}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 16 ) + 12}{2 \cdot 1}, x_{2} = \frac{- ( - 16 ) - 12}{2 \cdot 1}

x = \frac{- ( - 16 ) + 12}{2 \cdot 1} : 14

x = \frac{- ( - 16 ) - 12}{2 \cdot 1} : 2

The solutions to the quadratic equation are:

x = 14, x = 2

(2 x - 3)^{2} + x^{2} + 6 = (3 x + 1) (3 x - 1)

1 + 216 x^{9}

(\frac{1}{10} + \frac{1}{10})^{- 1}

(3 - 5 x) (1 - 2 x) \leq 0

x^{3} - 3 x + 2 > 0

What are the solutions to the equation x^2-16x+28=0 ?
The solutions to the equation x^2-16x+28=0 are x=14,x=2
Find the zeros of x^2-16x+28=0
The zeros of x^2-16x+28=0 are x=14,x=2