What are the solutions to the equation 2x^2-5+2=0 ?

The solutions to the equation 2x^2-5+2=0 are x=sqrt(3/2),x=-sqrt(3/2)

Find the zeros of 2x^2-5+2=0

The zeros of 2x^2-5+2=0 are x=sqrt(3/2),x=-sqrt(3/2)

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

x = \frac{3}{2}, x = - \frac{3}{2}

Decimal Notation

x = 1.22474 \dots, x = - 1.22474 \dots

Solution steps

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

Move

5

to the right side

2 x^{2} + 2 = 5

Move

2

to the right side

2 x^{2} = 3

Divide both sides by

2

x^{2} = \frac{3}{2}

For

x^{2} = f (a)

the solutions are

x = f (a), - f (a)

x = \frac{3}{2}, x = - \frac{3}{2}

x^{2} + 7 x - 98 = 0

x^{2} + 18 x - 4 = 0

a (x + b) - (x + a)^{2} = x^{2}

x^{2} - 16 x = - 55

6 x^{2} + 5 x = 4

What are the solutions to the equation 2x^2-5+2=0 ?
The solutions to the equation 2x^2-5+2=0 are x=sqrt(3/2),x=-sqrt(3/2)
Find the zeros of 2x^2-5+2=0
The zeros of 2x^2-5+2=0 are x=sqrt(3/2),x=-sqrt(3/2)