What are the solutions to the equation (X+2)(X-3)+2(3X-4)=3X+1 ?

The solutions to the equation (X+2)(X-3)+2(3X-4)=3X+1 are X=3,X=-5

Find the zeros of (X+2)(X-3)+2(3X-4)=3X+1

The zeros of (X+2)(X-3)+2(3X-4)=3X+1 are X=3,X=-5

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

(X + 2) (X - 3) + 2 (3 X - 4) = 3 X + 1

X = 3, X = - 5

Solution steps

(X + 2) (X - 3) + 2 (3 X - 4) = 3 X + 1

Expand

(X + 2) (X - 3) + 2 (3 X - 4) : X^{2} + 5 X - 14

X^{2} + 5 X - 14 = 3 X + 1

Move

1

to the left side

X^{2} + 5 X - 15 = 3 X

Move

3 X

to the left side

X^{2} + 2 X - 15 = 0

Solve with the quadratic formula

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

2^{2} - 4 \cdot 1 \cdot (- 15) = 8

X_{1, 2} = \frac{- 2 \pm 8}{2 \cdot 1}

Separate the solutions

X_{1} = \frac{- 2 + 8}{2 \cdot 1}, X_{2} = \frac{- 2 - 8}{2 \cdot 1}

X = \frac{- 2 + 8}{2 \cdot 1} : 3

X = \frac{- 2 - 8}{2 \cdot 1} : - 5

The solutions to the quadratic equation are:

X = 3, X = - 5

x^{2} + (\frac{90}{2000}) x + (\frac{1500}{2000}) = 0

x^{2} + (x - 7)^{2} = 132

(x + 13)^{2} = - 4

3 y^{2} + 2 y - 2 = 0

x, z = y^{2} - x^{2} + 2

What are the solutions to the equation (X+2)(X-3)+2(3X-4)=3X+1 ?
The solutions to the equation (X+2)(X-3)+2(3X-4)=3X+1 are X=3,X=-5
Find the zeros of (X+2)(X-3)+2(3X-4)=3X+1
The zeros of (X+2)(X-3)+2(3X-4)=3X+1 are X=3,X=-5