What are the solutions to the equation solvefor a,a^2+bc=1 ?

The solutions to the equation solvefor a,a^2+bc=1 are a=sqrt(1-bc),a=-sqrt(1-bc)

Find the zeros of solvefor a,a^2+bc=1

The zeros of solvefor a,a^2+bc=1 are a=sqrt(1-bc),a=-sqrt(1-bc)

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solve for

a, a^{2} + b c = 1

a = 1 - b c, a = - 1 - b c

Solution steps

a^{2} + b c = 1

Move

b c

to the right side

a^{2} = 1 - b c

For

x^{2} = f (a)

the solutions are

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

a = 1 - b c, a = - 1 - b c

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

- 5 \cdot t^{2} + 80 \cdot t = 0

- 12 x^{2} = - 32 x^{2}

12 x + 12 = 16 x^{2} - 16 x + 4

12 x^{2} = 16

What are the solutions to the equation solvefor a,a^2+bc=1 ?
The solutions to the equation solvefor a,a^2+bc=1 are a=sqrt(1-bc),a=-sqrt(1-bc)
Find the zeros of solvefor a,a^2+bc=1
The zeros of solvefor a,a^2+bc=1 are a=sqrt(1-bc),a=-sqrt(1-bc)