What are the solutions to the equation 4(x-2)^2+1=19 ?

The solutions to the equation 4(x-2)^2+1=19 are x= 3/(sqrt(2))+2,x=-3/(sqrt(2))+2

Find the zeros of 4(x-2)^2+1=19

The zeros of 4(x-2)^2+1=19 are x= 3/(sqrt(2))+2,x=-3/(sqrt(2))+2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

4 (x - 2)^{2} + 1 = 19

x = \frac{3}{2} + 2, x = - \frac{3}{2} + 2

Decimal Notation

x = 4.12132 \dots, x = - 0.12132 \dots

Solution steps

4 (x - 2)^{2} + 1 = 19

Move

1

to the right side

4 (x - 2)^{2} = 18

Divide both sides by

4

(x - 2)^{2} = \frac{9}{2}

For

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

the solutions are

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

Solve

x - 2 = \frac{9}{2} : x = \frac{3}{2} + 2

Solve

x - 2 = - \frac{9}{2} : x = - \frac{3}{2} + 2

The solutions to the quadratic equation are:

x = \frac{3}{2} + 2, x = - \frac{3}{2} + 2

(x + 1) (3 x - 1) - 4 = 0

x, x^{2} + \frac{y ^{2}}{2} = z^{2}

x^{2} - 4.5 x - 5 = 0

- 12 x^{2} + 11 x - 3 = 0

- 3 x^{2} + 8 x = \frac{( - 3 ) - ( - 3 )}{( 4 ) - ( 0 )}

What are the solutions to the equation 4(x-2)^2+1=19 ?
The solutions to the equation 4(x-2)^2+1=19 are x= 3/(sqrt(2))+2,x=-3/(sqrt(2))+2
Find the zeros of 4(x-2)^2+1=19
The zeros of 4(x-2)^2+1=19 are x= 3/(sqrt(2))+2,x=-3/(sqrt(2))+2