What are the solutions to the equation (x-5)(2x+3)=7 ?

The solutions to the equation (x-5)(2x+3)=7 are x= 11/2 ,x=-2

Find the zeros of (x-5)(2x+3)=7

The zeros of (x-5)(2x+3)=7 are x= 11/2 ,x=-2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

(x - 5) (2 x + 3) = 7

x = \frac{11}{2}, x = - 2

Decimal Notation

x = 5.5, x = - 2

Solution steps

(x - 5) (2 x + 3) = 7

Expand

(x - 5) (2 x + 3) : 2 x^{2} - 7 x - 15

2 x^{2} - 7 x - 15 = 7

Move

7

to the left side

2 x^{2} - 7 x - 22 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 7 ) \pm ( - 7 ) ^{2} - 4 \cdot 2 ( - 22 )}{2 \cdot 2}

(- 7)^{2} - 4 \cdot 2 (- 22) = 15

x_{1, 2} = \frac{- ( - 7 ) \pm 15}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- ( - 7 ) + 15}{2 \cdot 2}, x_{2} = \frac{- ( - 7 ) - 15}{2 \cdot 2}

x = \frac{- ( - 7 ) + 15}{2 \cdot 2} : \frac{11}{2}

x = \frac{- ( - 7 ) - 15}{2 \cdot 2} : - 2

The solutions to the quadratic equation are:

x = \frac{11}{2}, x = - 2

x^{2} - 1 x + 2 = 0

c (x) = x^{2} + 20 x + 900

(x) (x + 1) = 6 m^{2}

(x + 2) (x + 3) = 2

4 x^{2} + 22 x - 30 = 0

What are the solutions to the equation (x-5)(2x+3)=7 ?
The solutions to the equation (x-5)(2x+3)=7 are x= 11/2 ,x=-2
Find the zeros of (x-5)(2x+3)=7
The zeros of (x-5)(2x+3)=7 are x= 11/2 ,x=-2