What are the solutions to the equation x^2-4x=28 ?

The solutions to the equation x^2-4x=28 are x=2(1+2sqrt(2)),x=2(1-2sqrt(2))

Find the zeros of x^2-4x=28

The zeros of x^2-4x=28 are x=2(1+2sqrt(2)),x=2(1-2sqrt(2))

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x^{2} - 4 x = 28

x = 2 (1 + 22), x = 2 (1 - 22)

Decimal Notation

x = 7.65685 \dots, x = - 3.65685 \dots

Solution steps

x^{2} - 4 x = 28

Move

28

to the left side

x^{2} - 4 x - 28 = 0

Solve with the quadratic formula

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

(- 4)^{2} - 4 \cdot 1 \cdot (- 28) = 82

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

Separate the solutions

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

x = \frac{- ( - 4 ) + 8 2}{2 \cdot 1} : 2 (1 + 22)

x = \frac{- ( - 4 ) - 8 2}{2 \cdot 1} : 2 (1 - 22)

The solutions to the quadratic equation are:

x = 2 (1 + 22), x = 2 (1 - 22)

\frac{1 + \frac{9}{c - 1}}{1 - \frac{9}{c - 1}}

3 \cdot (\frac{1}{3})^{2}

(2 x + 1)^{2} = 0

(x^{- \frac{4}{7}})^{7}

What are the solutions to the equation x^2-4x=28 ?
The solutions to the equation x^2-4x=28 are x=2(1+2sqrt(2)),x=2(1-2sqrt(2))
Find the zeros of x^2-4x=28
The zeros of x^2-4x=28 are x=2(1+2sqrt(2)),x=2(1-2sqrt(2))