What are the solutions to the equation 4x^2+29=20x ?

The solutions to the equation 4x^2+29=20x are x= 5/2+i,x= 5/2-i

Find the zeros of 4x^2+29=20x

The zeros of 4x^2+29=20x are x= 5/2+i,x= 5/2-i

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

4 x^{2} + 29 = 20 x

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

Solution steps

4 x^{2} + 29 = 20 x

Move

20 x

to the left side

4 x^{2} + 29 - 20 x = 0

Write in the standard form

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

4 x^{2} - 20 x + 29 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 20 ) \pm ( - 20 ) ^{2} - 4 \cdot 4 \cdot 29}{2 \cdot 4}

Simplify

(- 20)^{2} - 4 \cdot 4 \cdot 29 : 8 i

x_{1, 2} = \frac{- ( - 20 ) \pm 8 i}{2 \cdot 4}

Separate the solutions

x_{1} = \frac{- ( - 20 ) + 8 i}{2 \cdot 4}, x_{2} = \frac{- ( - 20 ) - 8 i}{2 \cdot 4}

x = \frac{- ( - 20 ) + 8 i}{2 \cdot 4} : \frac{5}{2} + i

x = \frac{- ( - 20 ) - 8 i}{2 \cdot 4} : \frac{5}{2} - i

The solutions to the quadratic equation are:

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

10 a b^{7} + 2 a^{6} + 8 a^{4} b^{2} - 4 b

\frac{5 + 3}{4 - 5}

5 x - 3 = 3 + 2 x

\frac{2 x}{3} - \frac{2 x - 1}{4} = 2

2 x (x - 3) = - 4

What are the solutions to the equation 4x^2+29=20x ?
The solutions to the equation 4x^2+29=20x are x= 5/2+i,x= 5/2-i
Find the zeros of 4x^2+29=20x
The zeros of 4x^2+29=20x are x= 5/2+i,x= 5/2-i