What are the solutions to the equation x=2x^2-12x+25 ?

The solutions to the equation x=2x^2-12x+25 are x= 13/4+i(sqrt(31))/4 ,x= 13/4-i(sqrt(31))/4

Find the zeros of x=2x^2-12x+25

The zeros of x=2x^2-12x+25 are x= 13/4+i(sqrt(31))/4 ,x= 13/4-i(sqrt(31))/4

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

x = 2 x^{2} - 12 x + 25

x = \frac{13}{4} + i \frac{31}{4}, x = \frac{13}{4} - i \frac{31}{4}

Solution steps

x = 2 x^{2} - 12 x + 25

Switch sides

2 x^{2} - 12 x + 25 = x

Move

x

to the left side

2 x^{2} - 13 x + 25 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 13 ) \pm ( - 13 ) ^{2} - 4 \cdot 2 \cdot 25}{2 \cdot 2}

Simplify

(- 13)^{2} - 4 \cdot 2 \cdot 25 : 31 i

x_{1, 2} = \frac{- ( - 13 ) \pm 31 i}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- ( - 13 ) + 31 i}{2 \cdot 2}, x_{2} = \frac{- ( - 13 ) - 31 i}{2 \cdot 2}

x = \frac{- ( - 13 ) + 31 i}{2 \cdot 2} : \frac{13}{4} + i \frac{31}{4}

x = \frac{- ( - 13 ) - 31 i}{2 \cdot 2} : \frac{13}{4} - i \frac{31}{4}

The solutions to the quadratic equation are:

x = \frac{13}{4} + i \frac{31}{4}, x = \frac{13}{4} - i \frac{31}{4}

0.02 x^{2} + 2 x - 40 = 0

m^{2} - 8 m = 20

n^{2} + n - 930 = 0

(x - 3)^{2} = x (5 - x)

x^{2} + 23 + 3 = 0

What are the solutions to the equation x=2x^2-12x+25 ?
The solutions to the equation x=2x^2-12x+25 are x= 13/4+i(sqrt(31))/4 ,x= 13/4-i(sqrt(31))/4
Find the zeros of x=2x^2-12x+25
The zeros of x=2x^2-12x+25 are x= 13/4+i(sqrt(31))/4 ,x= 13/4-i(sqrt(31))/4