What are the solutions to the equation x^2+8x+4=28 ?

The solutions to the equation x^2+8x+4=28 are x=2(sqrt(10)-2),x=-2(2+sqrt(10))

Find the zeros of x^2+8x+4=28

The zeros of x^2+8x+4=28 are x=2(sqrt(10)-2),x=-2(2+sqrt(10))

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x^{2} + 8 x + 4 = 28

x = 2 (10 - 2), x = - 2 (2 + 10)

Decimal Notation

x = 2.32455 \dots, x = - 10.32455 \dots

Solution steps

x^{2} + 8 x + 4 = 28

Move

28

to the left side

x^{2} + 8 x - 24 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 8 \pm 8 ^{2} - 4 \cdot 1 \cdot ( - 24 )}{2 \cdot 1}

8^{2} - 4 \cdot 1 \cdot (- 24) = 410

x_{1, 2} = \frac{- 8 \pm 4 10}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 8 + 4 10}{2 \cdot 1}, x_{2} = \frac{- 8 - 4 10}{2 \cdot 1}

x = \frac{- 8 + 4 10}{2 \cdot 1} : 2 (10 - 2)

x = \frac{- 8 - 4 10}{2 \cdot 1} : - 2 (2 + 10)

The solutions to the quadratic equation are:

x = 2 (10 - 2), x = - 2 (2 + 10)

\frac{3}{5} + \frac{2}{3}

\frac{( - 2 m ^{4} ) ^{2}}{8 m ^{2}}

a^{\frac{2}{5}} \cdot a^{- 3}

\frac{1}{4} - \frac{1}{36}

3 - 108

What are the solutions to the equation x^2+8x+4=28 ?
The solutions to the equation x^2+8x+4=28 are x=2(sqrt(10)-2),x=-2(2+sqrt(10))
Find the zeros of x^2+8x+4=28
The zeros of x^2+8x+4=28 are x=2(sqrt(10)-2),x=-2(2+sqrt(10))