What are the solutions to the equation 2x^2+5=5x^2-37 ?

The solutions to the equation 2x^2+5=5x^2-37 are x=sqrt(14),x=-sqrt(14)

Find the zeros of 2x^2+5=5x^2-37

The zeros of 2x^2+5=5x^2-37 are x=sqrt(14),x=-sqrt(14)

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2 x^{2} + 5 = 5 x^{2} - 37

x = 14, x = - 14

Decimal Notation

x = 3.74165 \dots, x = - 3.74165 \dots

Solution steps

2 x^{2} + 5 = 5 x^{2} - 37

Move

5

to the right side

2 x^{2} = 5 x^{2} - 42

Move

5 x^{2}

to the left side

- 3 x^{2} = - 42

Divide both sides by

- 3

x^{2} = 14

For

x^{2} = f (a)

the solutions are

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

x = 14, x = - 14

(x + 3)^{2} - 3 = 49

4 (x^{2} + 6) + 35 x = 0

(x - 7)^{2} + (x + 1)^{2} = 7 (x - 1)^{2} + 43

12 x^{2} - 28 x + 12 = 0

x^{2} + 18 x = - 79

What are the solutions to the equation 2x^2+5=5x^2-37 ?
The solutions to the equation 2x^2+5=5x^2-37 are x=sqrt(14),x=-sqrt(14)
Find the zeros of 2x^2+5=5x^2-37
The zeros of 2x^2+5=5x^2-37 are x=sqrt(14),x=-sqrt(14)