What are the solutions to the equation ((2x^2+7))/3 =13 ?

The solutions to the equation ((2x^2+7))/3 =13 are x=4,x=-4

Find the zeros of ((2x^2+7))/3 =13

The zeros of ((2x^2+7))/3 =13 are x=4,x=-4

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\frac{( 2 x ^{2} + 7 )}{3} = 13

x = 4, x = - 4

Solution steps

\frac{( 2 x ^{2} + 7 )}{3} = 13

Multiply both sides by

3

2 x^{2} + 7 = 39

Move

7

to the right side

2 x^{2} = 32

Divide both sides by

2

x^{2} = 16

For

x^{2} = f (a)

the solutions are

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

x = 16, x = - 16

16 = 4

- 16 = - 4

x = 4, x = - 4

n^{2} - 2 n + 4 = 0

x^{2} - 4

15 + 3 x^{2} = 24

x^{2} - 2 x + 4

x^{2} - 2 x - 3

What are the solutions to the equation ((2x^2+7))/3 =13 ?
The solutions to the equation ((2x^2+7))/3 =13 are x=4,x=-4
Find the zeros of ((2x^2+7))/3 =13
The zeros of ((2x^2+7))/3 =13 are x=4,x=-4