What are the solutions to the equation 4+2(5x+2x^2)=11 ?

The solutions to the equation 4+2(5x+2x^2)=11 are x=(-5+sqrt(53))/4 ,x=-(5+sqrt(53))/4

Find the zeros of 4+2(5x+2x^2)=11

The zeros of 4+2(5x+2x^2)=11 are x=(-5+sqrt(53))/4 ,x=-(5+sqrt(53))/4

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

4 + 2 (5 x + 2 x^{2}) = 11

x = \frac{- 5 + 53}{4}, x = - \frac{5 + 53}{4}

Decimal Notation

x = 0.57002 \dots, x = - 3.07002 \dots

Solution steps

4 + 2 (5 x + 2 x^{2}) = 11

Expand

4 + 2 (5 x + 2 x^{2}) : 4 + 10 x + 4 x^{2}

4 + 10 x + 4 x^{2} = 11

Move

11

to the left side

4 x^{2} + 10 x - 7 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 10 \pm 1 0 ^{2} - 4 \cdot 4 ( - 7 )}{2 \cdot 4}

1 0^{2} - 4 \cdot 4 (- 7) = 253

x_{1, 2} = \frac{- 10 \pm 2 53}{2 \cdot 4}

Separate the solutions

x_{1} = \frac{- 10 + 2 53}{2 \cdot 4}, x_{2} = \frac{- 10 - 2 53}{2 \cdot 4}

x = \frac{- 10 + 2 53}{2 \cdot 4} : \frac{- 5 + 53}{4}

x = \frac{- 10 - 2 53}{2 \cdot 4} : - \frac{5 + 53}{4}

The solutions to the quadratic equation are:

x = \frac{- 5 + 53}{4}, x = - \frac{5 + 53}{4}

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

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

x^{2} - 5 x = 8

y, x^{2} + y^{2} = \frac{2 y}{x}

- 3 x^{2} + 12 x - 4 = - 12

What are the solutions to the equation 4+2(5x+2x^2)=11 ?
The solutions to the equation 4+2(5x+2x^2)=11 are x=(-5+sqrt(53))/4 ,x=-(5+sqrt(53))/4
Find the zeros of 4+2(5x+2x^2)=11
The zeros of 4+2(5x+2x^2)=11 are x=(-5+sqrt(53))/4 ,x=-(5+sqrt(53))/4