What are the solutions to the equation x^2+4x+53=0 ?

The solutions to the equation x^2+4x+53=0 are x=-2+7i,x=-2-7i

Find the zeros of x^2+4x+53=0

The zeros of x^2+4x+53=0 are x=-2+7i,x=-2-7i

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

x^{2} + 4 x + 53 = 0

x = - 2 + 7 i, x = - 2 - 7 i

Solution steps

x^{2} + 4 x + 53 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 4 \pm 4 ^{2} - 4 \cdot 1 \cdot 53}{2 \cdot 1}

Simplify

4^{2} - 4 \cdot 1 \cdot 53 : 14 i

x_{1, 2} = \frac{- 4 \pm 14 i}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 4 + 14 i}{2 \cdot 1}, x_{2} = \frac{- 4 - 14 i}{2 \cdot 1}

x = \frac{- 4 + 14 i}{2 \cdot 1} : - 2 + 7 i

x = \frac{- 4 - 14 i}{2 \cdot 1} : - 2 - 7 i

The solutions to the quadratic equation are:

x = - 2 + 7 i, x = - 2 - 7 i

- 8 \leq 10 - x \leq 4

y (y - 4) = 1

2 x^{3} - x^{2} (2 i - 1) - x (3 i + 5) + 2 + 2 i

7 m^{7} n - 28 m n^{3}

\frac{1 ^{2}}{2}

What are the solutions to the equation x^2+4x+53=0 ?
The solutions to the equation x^2+4x+53=0 are x=-2+7i,x=-2-7i
Find the zeros of x^2+4x+53=0
The zeros of x^2+4x+53=0 are x=-2+7i,x=-2-7i