What are the solutions to the equation 2x(x-14)+11(3+x)=-22x+27+x^2 ?

The solutions to the equation 2x(x-14)+11(3+x)=-22x+27+x^2 are x=-2,x=-3

Find the zeros of 2x(x-14)+11(3+x)=-22x+27+x^2

The zeros of 2x(x-14)+11(3+x)=-22x+27+x^2 are x=-2,x=-3

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

2 x (x - 14) + 11 (3 + x) = - 22 x + 27 + x^{2}

x = - 2, x = - 3

Solution steps

2 x (x - 14) + 11 (3 + x) = - 22 x + 27 + x^{2}

Expand

2 x (x - 14) + 11 (3 + x) : 2 x^{2} - 17 x + 33

2 x^{2} - 17 x + 33 = - 22 x + 27 + x^{2}

Move

x^{2}

to the left side

x^{2} - 17 x + 33 = - 22 x + 27

Move

27

to the left side

x^{2} - 17 x + 6 = - 22 x

Move

22 x

to the left side

x^{2} + 5 x + 6 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 5 \pm 5 ^{2} - 4 \cdot 1 \cdot 6}{2 \cdot 1}

5^{2} - 4 \cdot 1 \cdot 6 = 1

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

Separate the solutions

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

x = \frac{- 5 + 1}{2 \cdot 1} : - 2

x = \frac{- 5 - 1}{2 \cdot 1} : - 3

The solutions to the quadratic equation are:

x = - 2, x = - 3

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

t, y = v t + \frac{1}{2} g t^{2}

4 m^{2} + 200 m - 68696.06 = 0

2 a^{2} - 2 a - 3 = 0

2 x^{2} - x + 4

What are the solutions to the equation 2x(x-14)+11(3+x)=-22x+27+x^2 ?
The solutions to the equation 2x(x-14)+11(3+x)=-22x+27+x^2 are x=-2,x=-3
Find the zeros of 2x(x-14)+11(3+x)=-22x+27+x^2
The zeros of 2x(x-14)+11(3+x)=-22x+27+x^2 are x=-2,x=-3