What are the solutions to the equation (2k+3)^2-19=81 ?

The solutions to the equation (2k+3)^2-19=81 are k= 7/2 ,k=-13/2

Find the zeros of (2k+3)^2-19=81

The zeros of (2k+3)^2-19=81 are k= 7/2 ,k=-13/2

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(2 k + 3)^{2} - 19 = 81

k = \frac{7}{2}, k = - \frac{13}{2}

Decimal Notation

k = 3.5, k = - 6.5

Solution steps

(2 k + 3)^{2} - 19 = 81

Move

19

to the right side

(2 k + 3)^{2} = 100

For

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

the solutions are

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

Solve

2 k + 3 = 100 : k = \frac{7}{2}

Solve

2 k + 3 = - 100 : k = - \frac{13}{2}

The solutions to the quadratic equation are:

k = \frac{7}{2}, k = - \frac{13}{2}

100 + x^{2} = 220 + 2 x

x^{2} - 6 x + 9 = 0

12 x^{2} - 40 x + 25 = 0

x, x^{2} - y x + x - y + 2 = 0

18 = x^{2}

What are the solutions to the equation (2k+3)^2-19=81 ?
The solutions to the equation (2k+3)^2-19=81 are k= 7/2 ,k=-13/2
Find the zeros of (2k+3)^2-19=81
The zeros of (2k+3)^2-19=81 are k= 7/2 ,k=-13/2