What are the solutions to the equation (x)=4x^2-3x+25 ?

The solutions to the equation (x)=4x^2-3x+25 are x= 1/2+sqrt(6)i,x= 1/2-sqrt(6)i

Find the zeros of (x)=4x^2-3x+25

The zeros of (x)=4x^2-3x+25 are x= 1/2+sqrt(6)i,x= 1/2-sqrt(6)i

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

(x) = 4 x^{2} - 3 x + 25

x = \frac{1}{2} + 6 i, x = \frac{1}{2} - 6 i

Solution steps

(x) = 4 x^{2} - 3 x + 25

Refine

x = 4 x^{2} - 3 x + 25

Switch sides

4 x^{2} - 3 x + 25 = x

Move

x

to the left side

4 x^{2} - 4 x + 25 = 0

Solve with the quadratic formula

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

Simplify

(- 4)^{2} - 4 \cdot 4 \cdot 25 : 86 i

x_{1, 2} = \frac{- ( - 4 ) \pm 8 6 i}{2 \cdot 4}

Separate the solutions

x_{1} = \frac{- ( - 4 ) + 8 6 i}{2 \cdot 4}, x_{2} = \frac{- ( - 4 ) - 8 6 i}{2 \cdot 4}

x = \frac{- ( - 4 ) + 8 6 i}{2 \cdot 4} : \frac{1}{2} + 6 i

x = \frac{- ( - 4 ) - 8 6 i}{2 \cdot 4} : \frac{1}{2} - 6 i

The solutions to the quadratic equation are:

x = \frac{1}{2} + 6 i, x = \frac{1}{2} - 6 i

z^{2} + 6 z - 8 = 0

74 - 2 x^{2} = 84

9 x^{2} - 5 x (\frac{28 x}{33}) + 7 (\frac{28 x}{33})^{2} = 120

(x + 5) x = 0

(x^{2} - 4 x + 4) - \frac{( 20 x + 8 ) ^{2}}{225} = 4

What are the solutions to the equation (x)=4x^2-3x+25 ?
The solutions to the equation (x)=4x^2-3x+25 are x= 1/2+sqrt(6)i,x= 1/2-sqrt(6)i
Find the zeros of (x)=4x^2-3x+25
The zeros of (x)=4x^2-3x+25 are x= 1/2+sqrt(6)i,x= 1/2-sqrt(6)i