What are the solutions to the equation -2x^2+4x=3 ?

The solutions to the equation -2x^2+4x=3 are x=1-i(sqrt(2))/2 ,x=1+i(sqrt(2))/2

Find the zeros of -2x^2+4x=3

The zeros of -2x^2+4x=3 are x=1-i(sqrt(2))/2 ,x=1+i(sqrt(2))/2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

x = 1 - i \frac{2}{2}, x = 1 + i \frac{2}{2}

Solution steps

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

Move

3

to the left side

- 2 x^{2} + 4 x - 3 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 4 \pm 4 ^{2} - 4 ( - 2 ) ( - 3 )}{2 ( - 2 )}

Simplify

4^{2} - 4 (- 2) (- 3) : 22 i

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

Separate the solutions

x_{1} = \frac{- 4 + 2 2 i}{2 ( - 2 )}, x_{2} = \frac{- 4 - 2 2 i}{2 ( - 2 )}

x = \frac{- 4 + 2 2 i}{2 ( - 2 )} : 1 - i \frac{2}{2}

x = \frac{- 4 - 2 2 i}{2 ( - 2 )} : 1 + i \frac{2}{2}

The solutions to the quadratic equation are:

x = 1 - i \frac{2}{2}, x = 1 + i \frac{2}{2}

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

8 x^{2} + 18 x - 24 = 0

6 r^{2} - 7 r + 2 = 0

8 (6 - x) - 3 x (6 - x) = 0

y, y^{2} - x^{2} y + 3 x^{3} = 4

What are the solutions to the equation -2x^2+4x=3 ?
The solutions to the equation -2x^2+4x=3 are x=1-i(sqrt(2))/2 ,x=1+i(sqrt(2))/2
Find the zeros of -2x^2+4x=3
The zeros of -2x^2+4x=3 are x=1-i(sqrt(2))/2 ,x=1+i(sqrt(2))/2