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

The solutions to the equation 2x^2+4=5x are x= 5/4+i(sqrt(7))/4 ,x= 5/4-i(sqrt(7))/4

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

The zeros of 2x^2+4=5x are x= 5/4+i(sqrt(7))/4 ,x= 5/4-i(sqrt(7))/4

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

2 x^{2} + 4 = 5 x

x = \frac{5}{4} + i \frac{7}{4}, x = \frac{5}{4} - i \frac{7}{4}

Solution steps

2 x^{2} + 4 = 5 x

Move

5 x

to the left side

2 x^{2} + 4 - 5 x = 0

Write in the standard form

a x^{2} + b x + c = 0

2 x^{2} - 5 x + 4 = 0

Solve with the quadratic formula

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

Simplify

(- 5)^{2} - 4 \cdot 2 \cdot 4 : 7 i

x_{1, 2} = \frac{- ( - 5 ) \pm 7 i}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- ( - 5 ) + 7 i}{2 \cdot 2}, x_{2} = \frac{- ( - 5 ) - 7 i}{2 \cdot 2}

x = \frac{- ( - 5 ) + 7 i}{2 \cdot 2} : \frac{5}{4} + i \frac{7}{4}

x = \frac{- ( - 5 ) - 7 i}{2 \cdot 2} : \frac{5}{4} - i \frac{7}{4}

The solutions to the quadratic equation are:

x = \frac{5}{4} + i \frac{7}{4}, x = \frac{5}{4} - i \frac{7}{4}

12 x^{2} - 36 x + 10

g^{2} = 144

(x + 6)^{2} (x + 4) (x - 1) > 0

x^{2} - 8 x - 48 = 0

x = \frac{7.1 ( 95 )}{100 - ( 95 )}

What are the solutions to the equation 2x^2+4=5x ?
The solutions to the equation 2x^2+4=5x are x= 5/4+i(sqrt(7))/4 ,x= 5/4-i(sqrt(7))/4
Find the zeros of 2x^2+4=5x
The zeros of 2x^2+4=5x are x= 5/4+i(sqrt(7))/4 ,x= 5/4-i(sqrt(7))/4