What are the solutions to the equation 4x^2+4x+1=121 ?

The solutions to the equation 4x^2+4x+1=121 are x=5,x=-6

Find the zeros of 4x^2+4x+1=121

The zeros of 4x^2+4x+1=121 are x=5,x=-6

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

4 x^{2} + 4 x + 1 = 121

x = 5, x = - 6

Solution steps

4 x^{2} + 4 x + 1 = 121

Move

121

to the left side

4 x^{2} + 4 x - 120 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 4 \pm 4 ^{2} - 4 \cdot 4 ( - 120 )}{2 \cdot 4}

4^{2} - 4 \cdot 4 (- 120) = 44

x_{1, 2} = \frac{- 4 \pm 44}{2 \cdot 4}

Separate the solutions

x_{1} = \frac{- 4 + 44}{2 \cdot 4}, x_{2} = \frac{- 4 - 44}{2 \cdot 4}

x = \frac{- 4 + 44}{2 \cdot 4} : 5

x = \frac{- 4 - 44}{2 \cdot 4} : - 6

The solutions to the quadratic equation are:

x = 5, x = - 6

\frac{1}{2} (x - 4)^{2} + 3 = 30

4 x^{2} + 12 x + 6 = 0

x^{2} - 8 x + 15 = 0

x^{2} - x - 1 = 1

2 x - x^{2}

What are the solutions to the equation 4x^2+4x+1=121 ?
The solutions to the equation 4x^2+4x+1=121 are x=5,x=-6
Find the zeros of 4x^2+4x+1=121
The zeros of 4x^2+4x+1=121 are x=5,x=-6