What are the solutions to the equation x(x+1)=156 ?

The solutions to the equation x(x+1)=156 are x=12,x=-13

Find the zeros of x(x+1)=156

The zeros of x(x+1)=156 are x=12,x=-13

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

x (x + 1) = 156

x = 12, x = - 13

Solution steps

x (x + 1) = 156

Expand

x (x + 1) : x^{2} + x

x^{2} + x = 156

Move

156

to the left side

x^{2} + x - 156 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 \cdot 1 \cdot ( - 156 )}{2 \cdot 1}

1^{2} - 4 \cdot 1 \cdot (- 156) = 25

x_{1, 2} = \frac{- 1 \pm 25}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 1 + 25}{2 \cdot 1}, x_{2} = \frac{- 1 - 25}{2 \cdot 1}

x = \frac{- 1 + 25}{2 \cdot 1} : 12

x = \frac{- 1 - 25}{2 \cdot 1} : - 13

The solutions to the quadratic equation are:

x = 12, x = - 13

6 t^{2} + 7 t + 2 = 0

a, a^{2} - 2 ab + b^{2} + c^{2} = 0

x^{2} + (\frac{- 1}{2})^{2} = 1

- 8 x^{2} - 11 x - 2 = 0

\frac{( x + 2 ) ^{2}}{9} = \frac{2}{3} - \frac{( x + 3 ) ( x - 3 )}{5}

What are the solutions to the equation x(x+1)=156 ?
The solutions to the equation x(x+1)=156 are x=12,x=-13
Find the zeros of x(x+1)=156
The zeros of x(x+1)=156 are x=12,x=-13