What are the solutions to the equation 117^2=x(x+4) ?

The solutions to the equation 117^2=x(x+4) are x=-2+sqrt(13693),x=-2-sqrt(13693)

Find the zeros of 117^2=x(x+4)

The zeros of 117^2=x(x+4) are x=-2+sqrt(13693),x=-2-sqrt(13693)

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11 7^{2} = x (x + 4)

x = - 2 + 13693, x = - 2 - 13693

Decimal Notation

x = 115.01709 \dots, x = - 119.01709 \dots

Solution steps

11 7^{2} = x (x + 4)

Expand

11 7^{2} : 13689

Expand

x (x + 4) : x^{2} + 4 x

13689 = x^{2} + 4 x

Switch sides

x^{2} + 4 x = 13689

Move

13689

to the left side

x^{2} + 4 x - 13689 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 4 \pm 4 ^{2} - 4 \cdot 1 \cdot ( - 13689 )}{2 \cdot 1}

4^{2} - 4 \cdot 1 \cdot (- 13689) = 213693

x_{1, 2} = \frac{- 4 \pm 2 13693}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 4 + 2 13693}{2 \cdot 1}, x_{2} = \frac{- 4 - 2 13693}{2 \cdot 1}

x = \frac{- 4 + 2 13693}{2 \cdot 1} : - 2 + 13693

x = \frac{- 4 - 2 13693}{2 \cdot 1} : - 2 - 13693

The solutions to the quadratic equation are:

x = - 2 + 13693, x = - 2 - 13693

x + 1 + 4 x = 11 + 12 x - 3

\frac{6}{3}

\frac{( 2 m ^{3} n ^{4} ) ^{2} \cdot 2 m ^{2}}{m ^{3} n ^{4}}

9 + (\frac{8}{2})^{3}

(- 2)^{1}

What are the solutions to the equation 117^2=x(x+4) ?
The solutions to the equation 117^2=x(x+4) are x=-2+sqrt(13693),x=-2-sqrt(13693)
Find the zeros of 117^2=x(x+4)
The zeros of 117^2=x(x+4) are x=-2+sqrt(13693),x=-2-sqrt(13693)