What are the solutions to the equation 5(x-1)^2+7=127 ?

The solutions to the equation 5(x-1)^2+7=127 are x=2sqrt(6)+1,x=-2sqrt(6)+1

Find the zeros of 5(x-1)^2+7=127

The zeros of 5(x-1)^2+7=127 are x=2sqrt(6)+1,x=-2sqrt(6)+1

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5 (x - 1)^{2} + 7 = 127

x = 26 + 1, x = - 26 + 1

Decimal Notation

x = 5.89897 \dots, x = - 3.89897 \dots

Solution steps

5 (x - 1)^{2} + 7 = 127

Move

7

to the right side

5 (x - 1)^{2} = 120

Divide both sides by

5

(x - 1)^{2} = 24

For

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

the solutions are

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

Solve

x - 1 = 24 : x = 26 + 1

Solve

x - 1 = - 24 : x = - 26 + 1

The solutions to the quadratic equation are:

x = 26 + 1, x = - 26 + 1

4 - 72 x^{2} = 0

a, a^{2} b - b^{3} = 2 ab - 2 b^{2}

\frac{1}{2} x^{2} + 1 = 3

(30 + x) (35 + x) = 1300

0 = - x^{2} + 42 x - 20

What are the solutions to the equation 5(x-1)^2+7=127 ?
The solutions to the equation 5(x-1)^2+7=127 are x=2sqrt(6)+1,x=-2sqrt(6)+1
Find the zeros of 5(x-1)^2+7=127
The zeros of 5(x-1)^2+7=127 are x=2sqrt(6)+1,x=-2sqrt(6)+1