What are the solutions to the equation 4(x-7)^2-42=-22 ?

The solutions to the equation 4(x-7)^2-42=-22 are x=sqrt(5)+7,x=-sqrt(5)+7

Find the zeros of 4(x-7)^2-42=-22

The zeros of 4(x-7)^2-42=-22 are x=sqrt(5)+7,x=-sqrt(5)+7

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4 (x - 7)^{2} - 42 = - 22

x = 5 + 7, x = - 5 + 7

Decimal Notation

x = 9.23606 \dots, x = 4.76393 \dots

Solution steps

4 (x - 7)^{2} - 42 = - 22

Move

42

to the right side

4 (x - 7)^{2} = 20

Divide both sides by

4

(x - 7)^{2} = 5

For

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

the solutions are

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

Solve

x - 7 = 5 : x = 5 + 7

Solve

x - 7 = - 5 : x = - 5 + 7

The solutions to the quadratic equation are:

x = 5 + 7, x = - 5 + 7

93 < s - 31

(\frac{i}{3})^{3}

r^{2} + 3 = 0

V^{2} = 256

(x + 1) (x + 1)

What are the solutions to the equation 4(x-7)^2-42=-22 ?
The solutions to the equation 4(x-7)^2-42=-22 are x=sqrt(5)+7,x=-sqrt(5)+7
Find the zeros of 4(x-7)^2-42=-22
The zeros of 4(x-7)^2-42=-22 are x=sqrt(5)+7,x=-sqrt(5)+7