What are the solutions to the equation (2x+5)^2-7(2x+5)-30=0 ?

The solutions to the equation (2x+5)^2-7(2x+5)-30=0 are x= 5/2 ,x=-4

Find the zeros of (2x+5)^2-7(2x+5)-30=0

The zeros of (2x+5)^2-7(2x+5)-30=0 are x= 5/2 ,x=-4

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

(2 x + 5)^{2} - 7 (2 x + 5) - 30 = 0

x = \frac{5}{2}, x = - 4

Solution steps

(2 x + 5)^{2} - 7 (2 x + 5) - 30 = 0

Expand

(2 x + 5)^{2} - 7 (2 x + 5) - 30 : 4 x^{2} + 6 x - 40

4 x^{2} + 6 x - 40 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 6 \pm 6 ^{2} - 4 \cdot 4 ( - 40 )}{2 \cdot 4}

6^{2} - 4 \cdot 4 (- 40) = 26

x_{1, 2} = \frac{- 6 \pm 26}{2 \cdot 4}

Separate the solutions

x_{1} = \frac{- 6 + 26}{2 \cdot 4}, x_{2} = \frac{- 6 - 26}{2 \cdot 4}

x = \frac{- 6 + 26}{2 \cdot 4} : \frac{5}{2}

x = \frac{- 6 - 26}{2 \cdot 4} : - 4

The solutions to the quadratic equation are:

x = \frac{5}{2}, x = - 4

e^{i}

x^{2} + 8 x y + 15 y^{2}

x^{2} = 3 x + 3

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

- 4 (2 x + 5) + x + 2 = - 11

What are the solutions to the equation (2x+5)^2-7(2x+5)-30=0 ?
The solutions to the equation (2x+5)^2-7(2x+5)-30=0 are x= 5/2 ,x=-4
Find the zeros of (2x+5)^2-7(2x+5)-30=0
The zeros of (2x+5)^2-7(2x+5)-30=0 are x= 5/2 ,x=-4