What are the solutions to the equation 4x^2+20x=-13 ?

The solutions to the equation 4x^2+20x=-13 are x=(-5+2sqrt(3))/2 ,x=-(5+2sqrt(3))/2

Find the zeros of 4x^2+20x=-13

The zeros of 4x^2+20x=-13 are x=(-5+2sqrt(3))/2 ,x=-(5+2sqrt(3))/2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

4 x^{2} + 20 x = - 13

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

Decimal Notation

x = - 0.76794 \dots, x = - 4.23205 \dots

Solution steps

4 x^{2} + 20 x = - 13

Move

13

to the left side

4 x^{2} + 20 x + 13 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 20 \pm 2 0 ^{2} - 4 \cdot 4 \cdot 13}{2 \cdot 4}

2 0^{2} - 4 \cdot 4 \cdot 13 = 83

x_{1, 2} = \frac{- 20 \pm 8 3}{2 \cdot 4}

Separate the solutions

x_{1} = \frac{- 20 + 8 3}{2 \cdot 4}, x_{2} = \frac{- 20 - 8 3}{2 \cdot 4}

x = \frac{- 20 + 8 3}{2 \cdot 4} : \frac{- 5 + 2 3}{2}

x = \frac{- 20 - 8 3}{2 \cdot 4} : - \frac{5 + 2 3}{2}

The solutions to the quadratic equation are:

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

16 - (x + y)^{2}

- 13 = 9 r + 8

3 (4 + i) - 5 (- 3 + 6 i)

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

15 x (x - 5) < 3 (5 x - 1) (x - 6)

What are the solutions to the equation 4x^2+20x=-13 ?
The solutions to the equation 4x^2+20x=-13 are x=(-5+2sqrt(3))/2 ,x=-(5+2sqrt(3))/2
Find the zeros of 4x^2+20x=-13
The zeros of 4x^2+20x=-13 are x=(-5+2sqrt(3))/2 ,x=-(5+2sqrt(3))/2