What are the solutions to the equation 2*3+12=x^2-13 ?

The solutions to the equation 2*3+12=x^2-13 are x=sqrt(31),x=-sqrt(31)

Find the zeros of 2*3+12=x^2-13

The zeros of 2*3+12=x^2-13 are x=sqrt(31),x=-sqrt(31)

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2 \cdot 3 + 12 = x^{2} - 13

x = 31, x = - 31

Decimal Notation

x = 5.56776 \dots, x = - 5.56776 \dots

Solution steps

2 \cdot 3 + 12 = x^{2} - 13

Switch sides

x^{2} - 13 = 2 \cdot 3 + 12

Move

13

to the right side

x^{2} = 31

For

x^{2} = f (a)

the solutions are

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

x = 31, x = - 31

(2 x - 7) (x + 9) = 0

x, 3 y - 6 = 12 x^{2} - 6 y^{3}

9 x^{2} + 24 x - 24 = 0

6 x^{2} - 13 x - 28 = 0

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

What are the solutions to the equation 2*3+12=x^2-13 ?
The solutions to the equation 2*3+12=x^2-13 are x=sqrt(31),x=-sqrt(31)
Find the zeros of 2*3+12=x^2-13
The zeros of 2*3+12=x^2-13 are x=sqrt(31),x=-sqrt(31)