What are the solutions to the equation 30=(x-4)^2-5 ?

The solutions to the equation 30=(x-4)^2-5 are x=sqrt(35)+4,x=-sqrt(35)+4

Find the zeros of 30=(x-4)^2-5

The zeros of 30=(x-4)^2-5 are x=sqrt(35)+4,x=-sqrt(35)+4

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30 = (x - 4)^{2} - 5

x = 35 + 4, x = - 35 + 4

Decimal

x = 9.91607 \dots, x = - 1.91607 \dots

Solution steps

30 = (x - 4)^{2} - 5

Switch sides

(x - 4)^{2} - 5 = 30

Move

5

to the right side

(x - 4)^{2} = 35

For

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

the solutions are

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

Solve

x - 4 = 35 : x = 35 + 4

Solve

x - 4 = - 35 : x = - 35 + 4

The solutions to the quadratic equation are:

x = 35 + 4, x = - 35 + 4

110 + 3 x + 11 = 16 x + 4

4 x + 3 = 7

\frac{x}{x + 3} > 3

∣ 2 x + 1 ∣ > 5

\frac{7}{( 3 - x ) ( x + 3 )} > - 1

What are the solutions to the equation 30=(x-4)^2-5 ?
The solutions to the equation 30=(x-4)^2-5 are x=sqrt(35)+4,x=-sqrt(35)+4
Find the zeros of 30=(x-4)^2-5
The zeros of 30=(x-4)^2-5 are x=sqrt(35)+4,x=-sqrt(35)+4