What are the solutions to the equation 1100-q^2=300+q^2 ?

The solutions to the equation 1100-q^2=300+q^2 are q=20,q=-20

Find the zeros of 1100-q^2=300+q^2

The zeros of 1100-q^2=300+q^2 are q=20,q=-20

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1100 - q^{2} = 300 + q^{2}

q = 20, q = - 20

Solution steps

1100 - q^{2} = 300 + q^{2}

Move

1100

to the right side

- q^{2} = q^{2} - 800

Move

q^{2}

to the left side

- 2 q^{2} = - 800

Divide both sides by

- 2

q^{2} = 400

For

x^{2} = f (a)

the solutions are

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

q = 400, q = - 400

400 = 20

- 400 = - 20

q = 20, q = - 20

\frac{( x + 3 ) ( x + 1 )}{2} = 12

y^{2} = - 7 y

y^{2} = - 28

2 p^{2} + 8 p + (k + 1) = 0

3 x^{2} - 11 x - 42 = 0

What are the solutions to the equation 1100-q^2=300+q^2 ?
The solutions to the equation 1100-q^2=300+q^2 are q=20,q=-20
Find the zeros of 1100-q^2=300+q^2
The zeros of 1100-q^2=300+q^2 are q=20,q=-20