What are the solutions to the equation 4x^2-7=-4x ?

The solutions to the equation 4x^2-7=-4x are x=(-1+2sqrt(2))/2 ,x=-(1+2sqrt(2))/2

Find the zeros of 4x^2-7=-4x

The zeros of 4x^2-7=-4x are x=(-1+2sqrt(2))/2 ,x=-(1+2sqrt(2))/2

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4 x^{2} - 7 = - 4 x

x = \frac{- 1 + 2 2}{2}, x = - \frac{1 + 2 2}{2}

Decimal Notation

x = 0.91421 \dots, x = - 1.91421 \dots

Solution steps

4 x^{2} - 7 = - 4 x

Move

4 x

to the left side

4 x^{2} - 7 + 4 x = 0

Write in the standard form

a x^{2} + b x + c = 0

4 x^{2} + 4 x - 7 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 4 \pm 4 ^{2} - 4 \cdot 4 ( - 7 )}{2 \cdot 4}

4^{2} - 4 \cdot 4 (- 7) = 82

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

x = \frac{- 1 + 2 2}{2}, x = - \frac{1 + 2 2}{2}

3^{2} - 3

(- 3 - - 7)^{2}

39 - \frac{6}{25}

(2 x + 1)^{3}

\frac{x}{5} > 10

What are the solutions to the equation 4x^2-7=-4x ?
The solutions to the equation 4x^2-7=-4x are x=(-1+2sqrt(2))/2 ,x=-(1+2sqrt(2))/2
Find the zeros of 4x^2-7=-4x
The zeros of 4x^2-7=-4x are x=(-1+2sqrt(2))/2 ,x=-(1+2sqrt(2))/2