What are the solutions to the equation x^2=9-4x ?

The solutions to the equation x^2=9-4x are x=-2+sqrt(13),x=-2-sqrt(13)

Find the zeros of x^2=9-4x

The zeros of x^2=9-4x are x=-2+sqrt(13),x=-2-sqrt(13)

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x^{2} = 9 - 4 x

x = - 2 + 13, x = - 2 - 13

Decimal Notation

x = 1.60555 \dots, x = - 5.60555 \dots

Solution steps

x^{2} = 9 - 4 x

Move

4 x

to the left side

x^{2} + 4 x = 9

Move

9

to the left side

x^{2} + 4 x - 9 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 4 \pm 4 ^{2} - 4 \cdot 1 \cdot ( - 9 )}{2 \cdot 1}

4^{2} - 4 \cdot 1 \cdot (- 9) = 213

x_{1, 2} = \frac{- 4 \pm 2 13}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 4 + 2 13}{2 \cdot 1}, x_{2} = \frac{- 4 - 2 13}{2 \cdot 1}

x = \frac{- 4 + 2 13}{2 \cdot 1} : - 2 + 13

x = \frac{- 4 - 2 13}{2 \cdot 1} : - 2 - 13

The solutions to the quadratic equation are:

x = - 2 + 13, x = - 2 - 13

- \frac{3}{4} + \frac{7}{2}

0^{6}

\frac{x - 1}{x} \leq \frac{2 x}{x + 1} - \frac{x}{x - 1}

\frac{a}{2} \geq 3

\frac{1}{x + 1} + \frac{1}{x + 4} \leq 0

What are the solutions to the equation x^2=9-4x ?
The solutions to the equation x^2=9-4x are x=-2+sqrt(13),x=-2-sqrt(13)
Find the zeros of x^2=9-4x
The zeros of x^2=9-4x are x=-2+sqrt(13),x=-2-sqrt(13)