What are the solutions to the equation 5(x+3)^2-7=68 ?

The solutions to the equation 5(x+3)^2-7=68 are x=sqrt(15)-3,x=-sqrt(15)-3

Find the zeros of 5(x+3)^2-7=68

The zeros of 5(x+3)^2-7=68 are x=sqrt(15)-3,x=-sqrt(15)-3

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5 (x + 3)^{2} - 7 = 68

x = 15 - 3, x = - 15 - 3

Decimal Notation

x = 0.87298 \dots, x = - 6.87298 \dots

Solution steps

5 (x + 3)^{2} - 7 = 68

Move

7

to the right side

5 (x + 3)^{2} = 75

Divide both sides by

5

(x + 3)^{2} = 15

For

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

the solutions are

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

Solve

x + 3 = 15 : x = 15 - 3

Solve

x + 3 = - 15 : x = - 15 - 3

The solutions to the quadratic equation are:

x = 15 - 3, x = - 15 - 3

4 x (x - 1) = 3

t, f = 2 t - 3 f (t^{2})

y^{2} + y - 42 = 0

(x - 6) (x - 8) = - 1

- 3 x^{2} - 18 x - 26 = 0

What are the solutions to the equation 5(x+3)^2-7=68 ?
The solutions to the equation 5(x+3)^2-7=68 are x=sqrt(15)-3,x=-sqrt(15)-3
Find the zeros of 5(x+3)^2-7=68
The zeros of 5(x+3)^2-7=68 are x=sqrt(15)-3,x=-sqrt(15)-3