What are the solutions to the equation x^2+2(1)^2-2(1)-5=16 ?

The solutions to the equation x^2+2(1)^2-2(1)-5=16 are x=sqrt(21),x=-sqrt(21)

Find the zeros of x^2+2(1)^2-2(1)-5=16

The zeros of x^2+2(1)^2-2(1)-5=16 are x=sqrt(21),x=-sqrt(21)

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x^{2} + 2 (1)^{2} - 2 (1) - 5 = 16

x = 21, x = - 21

Decimal Notation

x = 4.58257 \dots, x = - 4.58257 \dots

Solution steps

x^{2} + 2 (1)^{2} - 2 (1) - 5 = 16

Expand

x^{2} + 2 \cdot 1^{2} - 2 (1) - 5 : x^{2} - 5

x^{2} - 5 = 16

Move

5

to the right side

x^{2} = 21

For

x^{2} = f (a)

the solutions are

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

x = 21, x = - 21

x^{2} - 6 x - 4 = 0

a^{2} + 1 = 3 a - 1

3 + 9 r^{2} = 12

s^{2} + 7 s + 1 = 0

- 0.003 x^{2} + 0.2 x + 3 = 0

What are the solutions to the equation x^2+2(1)^2-2(1)-5=16 ?
The solutions to the equation x^2+2(1)^2-2(1)-5=16 are x=sqrt(21),x=-sqrt(21)
Find the zeros of x^2+2(1)^2-2(1)-5=16
The zeros of x^2+2(1)^2-2(1)-5=16 are x=sqrt(21),x=-sqrt(21)