What are the solutions to the equation a=34x^2+x ?

The solutions to the equation a=34x^2+x are x=(-1+sqrt(1+136a))/(68),x=(-1-sqrt(1+136a))/(68)

Find the zeros of a=34x^2+x

The zeros of a=34x^2+x are x=(-1+sqrt(1+136a))/(68),x=(-1-sqrt(1+136a))/(68)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

a = 34 x^{2} + x

x = \frac{- 1 + 1 + 136 a}{68}, x = \frac{- 1 - 1 + 136 a}{68}

Solution steps

a = 34 x^{2} + x

Switch sides

34 x^{2} + x = a

Move

a

to the left side

34 x^{2} + x - a = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 \cdot 34 ( - a )}{2 \cdot 34}

Simplify

1^{2} - 4 \cdot 34 (- a) : 1 + 136 a

x_{1, 2} = \frac{- 1 \pm 1 + 136 a}{2 \cdot 34}

Separate the solutions

x_{1} = \frac{- 1 + 1 + 136 a}{2 \cdot 34}, x_{2} = \frac{- 1 - 1 + 136 a}{2 \cdot 34}

x = \frac{- 1 + 1 + 136 a}{2 \cdot 34} : \frac{- 1 + 1 + 136 a}{68}

x = \frac{- 1 - 1 + 136 a}{2 \cdot 34} : \frac{- 1 - 1 + 136 a}{68}

The solutions to the quadratic equation are:

x = \frac{- 1 + 1 + 136 a}{68}, x = \frac{- 1 - 1 + 136 a}{68}

r^{2} = (\frac{a}{2})^{2} + (\frac{a}{2})^{2}

- 14 = 5 n - n^{2}

2 x^{2} + 3 x + c = 0

a^{2} + 5 a + 4 = 0

x, x^{2} + x - p (p + 1) = 0

What are the solutions to the equation a=34x^2+x ?
The solutions to the equation a=34x^2+x are x=(-1+sqrt(1+136a))/(68),x=(-1-sqrt(1+136a))/(68)
Find the zeros of a=34x^2+x
The zeros of a=34x^2+x are x=(-1+sqrt(1+136a))/(68),x=(-1-sqrt(1+136a))/(68)