What are the solutions to the equation B=x^2+3x+1 ?

The solutions to the equation B=x^2+3x+1 are x=(-3+sqrt(4B+5))/2 ,x=(-3-sqrt(4B+5))/2

Find the zeros of B=x^2+3x+1

The zeros of B=x^2+3x+1 are x=(-3+sqrt(4B+5))/2 ,x=(-3-sqrt(4B+5))/2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

B = x^{2} + 3 x + 1

x = \frac{- 3 + 4 B + 5}{2}, x = \frac{- 3 - 4 B + 5}{2}

Solution steps

B = x^{2} + 3 x + 1

Switch sides

x^{2} + 3 x + 1 = B

Move

B

to the left side

x^{2} + 3 x + 1 - B = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 3 \pm 3 ^{2} - 4 \cdot 1 \cdot ( 1 - B )}{2 \cdot 1}

Simplify

3^{2} - 4 \cdot 1 \cdot (1 - B) : 4 B + 5

x_{1, 2} = \frac{- 3 \pm 4 B + 5}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 3 + 4 B + 5}{2 \cdot 1}, x_{2} = \frac{- 3 - 4 B + 5}{2 \cdot 1}

x = \frac{- 3 + 4 B + 5}{2 \cdot 1} : \frac{- 3 + 4 B + 5}{2}

x = \frac{- 3 - 4 B + 5}{2 \cdot 1} : \frac{- 3 - 4 B + 5}{2}

The solutions to the quadratic equation are:

x = \frac{- 3 + 4 B + 5}{2}, x = \frac{- 3 - 4 B + 5}{2}

x, z = x^{2} + y

l, T = 2 π l^{2}

x, 4 x^{2} + 4 y^{2} - 12 x + 27 = 0

25 = x^{2} + x + 4

x^{2} + 3 x - 5 = 0

What are the solutions to the equation B=x^2+3x+1 ?
The solutions to the equation B=x^2+3x+1 are x=(-3+sqrt(4B+5))/2 ,x=(-3-sqrt(4B+5))/2
Find the zeros of B=x^2+3x+1
The zeros of B=x^2+3x+1 are x=(-3+sqrt(4B+5))/2 ,x=(-3-sqrt(4B+5))/2