What are the solutions to the equation 4(x-2)^2+1=21 ?

The solutions to the equation 4(x-2)^2+1=21 are x=sqrt(5)+2,x=-sqrt(5)+2

Find the zeros of 4(x-2)^2+1=21

The zeros of 4(x-2)^2+1=21 are x=sqrt(5)+2,x=-sqrt(5)+2

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4 (x - 2)^{2} + 1 = 21

x = 5 + 2, x = - 5 + 2

Decimal Notation

x = 4.23606 \dots, x = - 0.23606 \dots

Solution steps

4 (x - 2)^{2} + 1 = 21

Move

1

to the right side

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

Divide both sides by

4

(x - 2)^{2} = 5

For

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

the solutions are

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

Solve

x - 2 = 5 : x = 5 + 2

Solve

x - 2 = - 5 : x = - 5 + 2

The solutions to the quadratic equation are:

x = 5 + 2, x = - 5 + 2

x^{2} - 2 (m - 2) x + (m^{2} - 2 m - 3) = 0

x^{2} + 16 x - 225 = 0

x, (p - 1) x^{2} + 2 p x + p - 1 = 0

m = x^{2} a + x^{2} b

2 x^{2} + 6 = - 1, 3 x = 0

What are the solutions to the equation 4(x-2)^2+1=21 ?
The solutions to the equation 4(x-2)^2+1=21 are x=sqrt(5)+2,x=-sqrt(5)+2
Find the zeros of 4(x-2)^2+1=21
The zeros of 4(x-2)^2+1=21 are x=sqrt(5)+2,x=-sqrt(5)+2