What are the solutions to the equation x(2x+5)=15 ?

The solutions to the equation x(2x+5)=15 are x=(-5+sqrt(145))/4 ,x=(-5-sqrt(145))/4

Find the zeros of x(2x+5)=15

The zeros of x(2x+5)=15 are x=(-5+sqrt(145))/4 ,x=(-5-sqrt(145))/4

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

x (2 x + 5) = 15

x = \frac{- 5 + 145}{4}, x = \frac{- 5 - 145}{4}

Decimal Notation

x = 1.76039 \dots, x = - 4.26039 \dots

Solution steps

x (2 x + 5) = 15

Expand

x (2 x + 5) : 2 x^{2} + 5 x

2 x^{2} + 5 x = 15

Move

15

to the left side

2 x^{2} + 5 x - 15 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 5 \pm 5 ^{2} - 4 \cdot 2 ( - 15 )}{2 \cdot 2}

5^{2} - 4 \cdot 2 (- 15) = 145

x_{1, 2} = \frac{- 5 \pm 145}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- 5 + 145}{2 \cdot 2}, x_{2} = \frac{- 5 - 145}{2 \cdot 2}

x = \frac{- 5 + 145}{2 \cdot 2} : \frac{- 5 + 145}{4}

x = \frac{- 5 - 145}{2 \cdot 2} : \frac{- 5 - 145}{4}

The solutions to the quadratic equation are:

x = \frac{- 5 + 145}{4}, x = \frac{- 5 - 145}{4}

x^{2} - 48 = 11

3 x^{2} - 3 = - 111

m^{2} - 4 m - 24 = 0

3 u^{2} - 10 u + 3 = 0

4 x^{2} + 32 x + 28 = 0

What are the solutions to the equation x(2x+5)=15 ?
The solutions to the equation x(2x+5)=15 are x=(-5+sqrt(145))/4 ,x=(-5-sqrt(145))/4
Find the zeros of x(2x+5)=15
The zeros of x(2x+5)=15 are x=(-5+sqrt(145))/4 ,x=(-5-sqrt(145))/4