What are the solutions to the equation k=r^2+5,6<= r<= 7 ?

The solutions to the equation k=r^2+5,6<= r<= 7 are r=sqrt(k-5),r=-sqrt(k-5)

Find the zeros of k=r^2+5,6<= r<= 7

The zeros of k=r^2+5,6<= r<= 7 are r=sqrt(k-5),r=-sqrt(k-5)

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The solutions to the equation k=r^2+5,6<= r<= 7 are r=sqrt(k-5),r=-sqrt(k-5)

The zeros of k=r^2+5,6<= r<= 7 are r=sqrt(k-5),r=-sqrt(k-5)

k = r^{2} + 5, 6 \leq r \leq 7

r = k - 5, r = - k - 5

Solution steps

k = r^{2} + 5, 6 \leq r \leq 7

Switch sides

r^{2} + 5 = k

Move

5

to the right side

r^{2} = k - 5

For

x^{2} = f (a)

the solutions are

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

r = k - 5, r = - k - 5

1 x^{2} + 7 x + 15 = 0

3 x^{2} + x

x^{2} + 3 x - 24 = 3 x + 13 x + 1

x^{2} - 15 x - 4 = 0

6 x^{2} - 30 x - 36 = 56

What are the solutions to the equation k=r^2+5,6<= r<= 7 ?
The solutions to the equation k=r^2+5,6<= r<= 7 are r=sqrt(k-5),r=-sqrt(k-5)
Find the zeros of k=r^2+5,6<= r<= 7
The zeros of k=r^2+5,6<= r<= 7 are r=sqrt(k-5),r=-sqrt(k-5)