What are the solutions to the equation 4+5x-2x^2=5 ?

The solutions to the equation 4+5x-2x^2=5 are x=(5-sqrt(17))/4 ,x=(5+sqrt(17))/4

Find the zeros of 4+5x-2x^2=5

The zeros of 4+5x-2x^2=5 are x=(5-sqrt(17))/4 ,x=(5+sqrt(17))/4

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

x = \frac{5 - 17}{4}, x = \frac{5 + 17}{4}

Solution steps

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

Move

5

to the left side

- 2 x^{2} + 5 x - 1 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 5 \pm 5 ^{2} - 4 ( - 2 ) ( - 1 )}{2 ( - 2 )}

5^{2} - 4 (- 2) (- 1) = 17

x_{1, 2} = \frac{- 5 \pm 17}{2 ( - 2 )}

Separate the solutions

x_{1} = \frac{- 5 + 17}{2 ( - 2 )}, x_{2} = \frac{- 5 - 17}{2 ( - 2 )}

x = \frac{- 5 + 17}{2 ( - 2 )} : \frac{5 - 17}{4}

x = \frac{- 5 - 17}{2 ( - 2 )} : \frac{5 + 17}{4}

The solutions to the quadratic equation are:

x = \frac{5 - 17}{4}, x = \frac{5 + 17}{4}

\frac{x - 5}{3} + \frac{x + 4}{2} \geq \frac{x + 3}{6}

(x^{4} y^{2} z) (y^{2} z)

(x + 4 a)^{2} - 25 a^{2}

m^{2} + 21 m - 22 = 0

lo g_{4} (2) + lo g_{4} (32)

What are the solutions to the equation 4+5x-2x^2=5 ?
The solutions to the equation 4+5x-2x^2=5 are x=(5-sqrt(17))/4 ,x=(5+sqrt(17))/4
Find the zeros of 4+5x-2x^2=5
The zeros of 4+5x-2x^2=5 are x=(5-sqrt(17))/4 ,x=(5+sqrt(17))/4