What are the solutions to the equation (c+1)^2= 21/2 ?

The solutions to the equation (c+1)^2= 21/2 are c=sqrt(21/2)-1,c=-sqrt(21/2)-1

Find the zeros of (c+1)^2= 21/2

The zeros of (c+1)^2= 21/2 are c=sqrt(21/2)-1,c=-sqrt(21/2)-1

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(c + 1)^{2} = \frac{21}{2}

c = \frac{21}{2} - 1, c = - \frac{21}{2} - 1

Decimal Notation

c = 2.24037 \dots, c = - 4.24037 \dots

Solution steps

(c + 1)^{2} = \frac{21}{2}

For

(g (x))^{2} = f (a)

the solutions are

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

Solve

c + 1 = \frac{21}{2} : c = \frac{21}{2} - 1

Solve

c + 1 = - \frac{21}{2} : c = - \frac{21}{2} - 1

The solutions to the quadratic equation are:

c = \frac{21}{2} - 1, c = - \frac{21}{2} - 1

x^{2} + 8 = - 6 x

- 5 (t^{2} - 3) + 15 = - 2 t

x^{2} + 6 x - 21 = 0

x^{2} = 15 - 2 x

- x + 4 x (2 - x) = - 1

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