What are the solutions to the equation x^2-12x+28=0 ?

The solutions to the equation x^2-12x+28=0 are x=2(3+sqrt(2)),x=2(3-sqrt(2))

Find the zeros of x^2-12x+28=0

The zeros of x^2-12x+28=0 are x=2(3+sqrt(2)),x=2(3-sqrt(2))

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x^{2} - 12 x + 28 = 0

x = 2 (3 + 2), x = 2 (3 - 2)

Decimal Notation

x = 8.82842 \dots, x = 3.17157 \dots

Solution steps

x^{2} - 12 x + 28 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 12 ) \pm ( - 12 ) ^{2} - 4 \cdot 1 \cdot 28}{2 \cdot 1}

(- 12)^{2} - 4 \cdot 1 \cdot 28 = 42

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

Separate the solutions

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

x = \frac{- ( - 12 ) + 4 2}{2 \cdot 1} : 2 (3 + 2)

x = \frac{- ( - 12 ) - 4 2}{2 \cdot 1} : 2 (3 - 2)

The solutions to the quadratic equation are:

x = 2 (3 + 2), x = 2 (3 - 2)

∣ 5 - 2 x ∣ = 11

\frac{1}{x + 1} + \frac{1}{x + 8} \leq \frac{1}{x + 9}

14 r - 11 > - 227

5108

(x - 5) (x + 2)^{2} > 0

What are the solutions to the equation x^2-12x+28=0 ?
The solutions to the equation x^2-12x+28=0 are x=2(3+sqrt(2)),x=2(3-sqrt(2))
Find the zeros of x^2-12x+28=0
The zeros of x^2-12x+28=0 are x=2(3+sqrt(2)),x=2(3-sqrt(2))