What are the solutions to the equation (2x+5)^2=9x ?

The solutions to the equation (2x+5)^2=9x are x=-11/8+i(3sqrt(31))/8 ,x=-11/8-i(3sqrt(31))/8

Find the zeros of (2x+5)^2=9x

The zeros of (2x+5)^2=9x are x=-11/8+i(3sqrt(31))/8 ,x=-11/8-i(3sqrt(31))/8

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

(2 x + 5)^{2} = 9 x

x = - \frac{11}{8} + i \frac{3 31}{8}, x = - \frac{11}{8} - i \frac{3 31}{8}

Solution steps

(2 x + 5)^{2} = 9 x

Expand

(2 x + 5)^{2} : 4 x^{2} + 20 x + 25

4 x^{2} + 20 x + 25 = 9 x

Move

9 x

to the left side

4 x^{2} + 11 x + 25 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 11 \pm 1 1 ^{2} - 4 \cdot 4 \cdot 25}{2 \cdot 4}

Simplify

1 1^{2} - 4 \cdot 4 \cdot 25 : 331 i

x_{1, 2} = \frac{- 11 \pm 3 31 i}{2 \cdot 4}

Separate the solutions

x_{1} = \frac{- 11 + 3 31 i}{2 \cdot 4}, x_{2} = \frac{- 11 - 3 31 i}{2 \cdot 4}

x = \frac{- 11 + 3 31 i}{2 \cdot 4} : - \frac{11}{8} + i \frac{3 31}{8}

x = \frac{- 11 - 3 31 i}{2 \cdot 4} : - \frac{11}{8} - i \frac{3 31}{8}

The solutions to the quadratic equation are:

x = - \frac{11}{8} + i \frac{3 31}{8}, x = - \frac{11}{8} - i \frac{3 31}{8}

3 x^{2} - 7 x + 12 = 0

2 x^{2} + 1 < 9 x - 3

4 + 5 x = 36 - 3 x

t^{- 1}

What are the solutions to the equation (2x+5)^2=9x ?
The solutions to the equation (2x+5)^2=9x are x=-11/8+i(3sqrt(31))/8 ,x=-11/8-i(3sqrt(31))/8
Find the zeros of (2x+5)^2=9x
The zeros of (2x+5)^2=9x are x=-11/8+i(3sqrt(31))/8 ,x=-11/8-i(3sqrt(31))/8