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

The solutions to the equation 4x^2=4-x are x=(-1+sqrt(65))/8 ,x=(-1-sqrt(65))/8

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

The zeros of 4x^2=4-x are x=(-1+sqrt(65))/8 ,x=(-1-sqrt(65))/8

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4 x^{2} = 4 - x

x = \frac{- 1 + 65}{8}, x = \frac{- 1 - 65}{8}

Decimal Notation

x = 0.88278 \dots, x = - 1.13278 \dots

Solution steps

4 x^{2} = 4 - x

Move

x

to the left side

4 x^{2} + x = 4

Move

4

to the left side

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

Solve with the quadratic formula

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

1^{2} - 4 \cdot 4 (- 4) = 65

x_{1, 2} = \frac{- 1 \pm 65}{2 \cdot 4}

Separate the solutions

x_{1} = \frac{- 1 + 65}{2 \cdot 4}, x_{2} = \frac{- 1 - 65}{2 \cdot 4}

x = \frac{- 1 + 65}{2 \cdot 4} : \frac{- 1 + 65}{8}

x = \frac{- 1 - 65}{2 \cdot 4} : \frac{- 1 - 65}{8}

The solutions to the quadratic equation are:

x = \frac{- 1 + 65}{8}, x = \frac{- 1 - 65}{8}

x = lo g_{3} (9)

2 (3 a^{2} - 1) - 6 a = 5 a

(2 + i)^{2} - (4 - i)^{2}

x - 4 \geq 1

(x - 4) (3 x + 2) = 0

What are the solutions to the equation 4x^2=4-x ?
The solutions to the equation 4x^2=4-x are x=(-1+sqrt(65))/8 ,x=(-1-sqrt(65))/8
Find the zeros of 4x^2=4-x
The zeros of 4x^2=4-x are x=(-1+sqrt(65))/8 ,x=(-1-sqrt(65))/8