What are the solutions to the equation 5x^2+6x+5=2(2x^2+10)-4x ?

The solutions to the equation 5x^2+6x+5=2(2x^2+10)-4x are x=-5+2sqrt(10),x=-5-2sqrt(10)

Find the zeros of 5x^2+6x+5=2(2x^2+10)-4x

The zeros of 5x^2+6x+5=2(2x^2+10)-4x are x=-5+2sqrt(10),x=-5-2sqrt(10)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

5 x^{2} + 6 x + 5 = 2 (2 x^{2} + 10) - 4 x

x = - 5 + 210, x = - 5 - 210

Decimal Notation

x = 1.32455 \dots, x = - 11.32455 \dots

Solution steps

5 x^{2} + 6 x + 5 = 2 (2 x^{2} + 10) - 4 x

Expand

2 (2 x^{2} + 10) - 4 x : 4 x^{2} + 20 - 4 x

5 x^{2} + 6 x + 5 = 4 x^{2} + 20 - 4 x

Move

4 x

to the left side

5 x^{2} + 10 x + 5 = 4 x^{2} + 20

Move

20

to the left side

5 x^{2} + 10 x - 15 = 4 x^{2}

Move

4 x^{2}

to the left side

x^{2} + 10 x - 15 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 10 \pm 1 0 ^{2} - 4 \cdot 1 \cdot ( - 15 )}{2 \cdot 1}

1 0^{2} - 4 \cdot 1 \cdot (- 15) = 410

x_{1, 2} = \frac{- 10 \pm 4 10}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 10 + 4 10}{2 \cdot 1}, x_{2} = \frac{- 10 - 4 10}{2 \cdot 1}

x = \frac{- 10 + 4 10}{2 \cdot 1} : - 5 + 210

x = \frac{- 10 - 4 10}{2 \cdot 1} : - 5 - 210

The solutions to the quadratic equation are:

x = - 5 + 210, x = - 5 - 210

0 = 19.23 x + 3.25 x^{2}

- \frac{\frac{8}{7}}{( - \frac{10}{9} )}

x^{2} < x + 6

m^{2} - 6 m + 8 > 0

m^{2} - 2 m - 168

What are the solutions to the equation 5x^2+6x+5=2(2x^2+10)-4x ?
The solutions to the equation 5x^2+6x+5=2(2x^2+10)-4x are x=-5+2sqrt(10),x=-5-2sqrt(10)
Find the zeros of 5x^2+6x+5=2(2x^2+10)-4x
The zeros of 5x^2+6x+5=2(2x^2+10)-4x are x=-5+2sqrt(10),x=-5-2sqrt(10)