What are the solutions to the equation 5x^2=4-2x ?

The solutions to the equation 5x^2=4-2x are x=(-1+sqrt(21))/5 ,x=-(1+sqrt(21))/5

Find the zeros of 5x^2=4-2x

The zeros of 5x^2=4-2x are x=(-1+sqrt(21))/5 ,x=-(1+sqrt(21))/5

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5 x^{2} = 4 - 2 x

x = \frac{- 1 + 21}{5}, x = - \frac{1 + 21}{5}

Decimal Notation

x = 0.71651 \dots, x = - 1.11651 \dots

Solution steps

5 x^{2} = 4 - 2 x

Move

2 x

to the left side

5 x^{2} + 2 x = 4

Move

4

to the left side

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

Solve with the quadratic formula

x_{1, 2} = \frac{- 2 \pm 2 ^{2} - 4 \cdot 5 ( - 4 )}{2 \cdot 5}

2^{2} - 4 \cdot 5 (- 4) = 221

x_{1, 2} = \frac{- 2 \pm 2 21}{2 \cdot 5}

Separate the solutions

x_{1} = \frac{- 2 + 2 21}{2 \cdot 5}, x_{2} = \frac{- 2 - 2 21}{2 \cdot 5}

x = \frac{- 2 + 2 21}{2 \cdot 5} : \frac{- 1 + 21}{5}

x = \frac{- 2 - 2 21}{2 \cdot 5} : - \frac{1 + 21}{5}

The solutions to the quadratic equation are:

x = \frac{- 1 + 21}{5}, x = - \frac{1 + 21}{5}

- 4 = - 4 x^{2} - 6 x

2 x^{2} - \frac{1}{32} = 0

t, 7 = 16 - 9.8 t^{2}

x^{2} + 3 x - 15 = - 4

2 x (x - 2) = x + 3

What are the solutions to the equation 5x^2=4-2x ?
The solutions to the equation 5x^2=4-2x are x=(-1+sqrt(21))/5 ,x=-(1+sqrt(21))/5
Find the zeros of 5x^2=4-2x
The zeros of 5x^2=4-2x are x=(-1+sqrt(21))/5 ,x=-(1+sqrt(21))/5