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 \cdot 2 \cdot 3}{2 \cdot 2}

Simplify

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

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

Separate the solutions

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

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

x = \frac{- ( - 4 ) - 2 2 i}{2 \cdot 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}

z^{2} - 9 = 0

5^{2} + 9^{2} = x^{2}

x, x^{2} + 4 x + 1 = 0

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

3 x^{2} - 12 = - 16 x

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