What are the solutions to the equation 5x^2+9=4x+5 ?

The solutions to the equation 5x^2+9=4x+5 are x= 2/5+i 4/5 ,x= 2/5-i 4/5

Find the zeros of 5x^2+9=4x+5

The zeros of 5x^2+9=4x+5 are x= 2/5+i 4/5 ,x= 2/5-i 4/5

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

5 x^{2} + 9 = 4 x + 5

x = \frac{2}{5} + i \frac{4}{5}, x = \frac{2}{5} - i \frac{4}{5}

Solution steps

5 x^{2} + 9 = 4 x + 5

Move

5

to the left side

5 x^{2} + 4 = 4 x

Move

4 x

to the left side

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

Write in the standard form

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

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

Solve with the quadratic formula

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

Simplify

(- 4)^{2} - 4 \cdot 5 \cdot 4 : 8 i

x_{1, 2} = \frac{- ( - 4 ) \pm 8 i}{2 \cdot 5}

Separate the solutions

x_{1} = \frac{- ( - 4 ) + 8 i}{2 \cdot 5}, x_{2} = \frac{- ( - 4 ) - 8 i}{2 \cdot 5}

x = \frac{- ( - 4 ) + 8 i}{2 \cdot 5} : \frac{2}{5} + i \frac{4}{5}

x = \frac{- ( - 4 ) - 8 i}{2 \cdot 5} : \frac{2}{5} - i \frac{4}{5}

The solutions to the quadratic equation are:

x = \frac{2}{5} + i \frac{4}{5}, x = \frac{2}{5} - i \frac{4}{5}

3 (k - 2) - 6 = 4 k - (3 k - 1)

- (\frac{2}{3})^{3}

4 x + 6 + x < 4 + 3 x - 4

\frac{x ^{5} y ^{2}}{x y ^{2}}

(x + 3) (x + 4) \geq 0

What are the solutions to the equation 5x^2+9=4x+5 ?
The solutions to the equation 5x^2+9=4x+5 are x= 2/5+i 4/5 ,x= 2/5-i 4/5
Find the zeros of 5x^2+9=4x+5
The zeros of 5x^2+9=4x+5 are x= 2/5+i 4/5 ,x= 2/5-i 4/5