What are the solutions to the equation x^2-12x+4=0 ?

The solutions to the equation x^2-12x+4=0 are x=2(3+2sqrt(2)),x=2(3-2sqrt(2))

Find the zeros of x^2-12x+4=0

The zeros of x^2-12x+4=0 are x=2(3+2sqrt(2)),x=2(3-2sqrt(2))

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x^{2} - 12 x + 4 = 0

x = 2 (3 + 22), x = 2 (3 - 22)

Decimal Notation

x = 11.65685 \dots, x = 0.34314 \dots

Solution steps

x^{2} - 12 x + 4 = 0

Solve with the quadratic formula

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

(- 12)^{2} - 4 \cdot 1 \cdot 4 = 82

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

Separate the solutions

x_{1} = \frac{- ( - 12 ) + 8 2}{2 \cdot 1}, x_{2} = \frac{- ( - 12 ) - 8 2}{2 \cdot 1}

x = \frac{- ( - 12 ) + 8 2}{2 \cdot 1} : 2 (3 + 22)

x = \frac{- ( - 12 ) - 8 2}{2 \cdot 1} : 2 (3 - 22)

The solutions to the quadratic equation are:

x = 2 (3 + 22), x = 2 (3 - 22)

(- 3 + i)^{6}

\frac{x + 1}{( x + 2 ) ( x - 2 )} > 0

\frac{x ^{3}}{x ^{9}}

8 x^{2} + 28 x + 12

What are the solutions to the equation x^2-12x+4=0 ?
The solutions to the equation x^2-12x+4=0 are x=2(3+2sqrt(2)),x=2(3-2sqrt(2))
Find the zeros of x^2-12x+4=0
The zeros of x^2-12x+4=0 are x=2(3+2sqrt(2)),x=2(3-2sqrt(2))