What are the solutions to the equation 5/2 x^2+2/5 =1-2x ?

The solutions to the equation 5/2 x^2+2/5 =1-2x are x=(-2+sqrt(10))/5 ,x=-(2+sqrt(10))/5

Find the zeros of 5/2 x^2+2/5 =1-2x

The zeros of 5/2 x^2+2/5 =1-2x are x=(-2+sqrt(10))/5 ,x=-(2+sqrt(10))/5

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\frac{5}{2} x^{2} + \frac{2}{5} = 1 - 2 x

x = \frac{- 2 + 10}{5}, x = - \frac{2 + 10}{5}

Decimal Notation

x = 0.23245 \dots, x = - 1.03245 \dots

Solution steps

\frac{5}{2} x^{2} + \frac{2}{5} = 1 - 2 x

Find Least Common Multiplier of

2, 5 : 10

Multiply by LCM=

10

\frac{5}{2} x^{2} \cdot 10 + \frac{2}{5} \cdot 10 = 1 \cdot 10 - 2 x \cdot 10

Simplify

25 x^{2} + 4 = 10 - 20 x

Move

20 x

to the left side

25 x^{2} + 4 + 20 x = 10

Move

10

to the left side

25 x^{2} + 20 x - 6 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 20 \pm 2 0 ^{2} - 4 \cdot 25 ( - 6 )}{2 \cdot 25}

2 0^{2} - 4 \cdot 25 (- 6) = 1010

x_{1, 2} = \frac{- 20 \pm 10 10}{2 \cdot 25}

Separate the solutions

x_{1} = \frac{- 20 + 10 10}{2 \cdot 25}, x_{2} = \frac{- 20 - 10 10}{2 \cdot 25}

x = \frac{- 20 + 10 10}{2 \cdot 25} : \frac{- 2 + 10}{5}

x = \frac{- 20 - 10 10}{2 \cdot 25} : - \frac{2 + 10}{5}

The solutions to the quadratic equation are:

x = \frac{- 2 + 10}{5}, x = - \frac{2 + 10}{5}

x, f (x + 1) = 3 x^{2} + 4 x - 3

x, x^{2} = t^{2} + 8

x, (p - 2) \cdot x^{2} + 3 x + 5 = 0

3 x^{2} + 5 x - 1300 = 0

d = (x + 4)^{2} + (2 x + 1) (2 x - 1) - 3 x (\frac{x - 1}{3})

What are the solutions to the equation 5/2 x^2+2/5 =1-2x ?
The solutions to the equation 5/2 x^2+2/5 =1-2x are x=(-2+sqrt(10))/5 ,x=-(2+sqrt(10))/5
Find the zeros of 5/2 x^2+2/5 =1-2x
The zeros of 5/2 x^2+2/5 =1-2x are x=(-2+sqrt(10))/5 ,x=-(2+sqrt(10))/5