What are the solutions to the equation (7x+4)^2=81 ?

The solutions to the equation (7x+4)^2=81 are x= 5/7 ,x=-13/7

Find the zeros of (7x+4)^2=81

The zeros of (7x+4)^2=81 are x= 5/7 ,x=-13/7

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

(7 x + 4)^{2} = 81

x = \frac{5}{7}, x = - \frac{13}{7}

Decimal Notation

x = 0.71428 \dots, x = - 1.85714 \dots

Solution steps

(7 x + 4)^{2} = 81

For

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

the solutions are

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

Solve

7 x + 4 = 81 : x = \frac{5}{7}

Solve

7 x + 4 = - 81 : x = - \frac{13}{7}

The solutions to the quadratic equation are:

x = \frac{5}{7}, x = - \frac{13}{7}

\frac{15 c ^{2} - 8 c}{3} = 4

4 x^{2} + 2 x + 5

4.9 x^{2} - 90 x - 150 = 0

5 x^{2} + 31623 x + 50 = 0

- 4.9 t^{2} + 18 t + 7.4 = 0

What are the solutions to the equation (7x+4)^2=81 ?
The solutions to the equation (7x+4)^2=81 are x= 5/7 ,x=-13/7
Find the zeros of (7x+4)^2=81
The zeros of (7x+4)^2=81 are x= 5/7 ,x=-13/7