What are the solutions to the equation 6x^2-4x+7=4x^2+5x ?

The solutions to the equation 6x^2-4x+7=4x^2+5x are x= 7/2 ,x=1

Find the zeros of 6x^2-4x+7=4x^2+5x

The zeros of 6x^2-4x+7=4x^2+5x are x= 7/2 ,x=1

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

6 x^{2} - 4 x + 7 = 4 x^{2} + 5 x

x = \frac{7}{2}, x = 1

Solution steps

6 x^{2} - 4 x + 7 = 4 x^{2} + 5 x

Move

5 x

to the left side

6 x^{2} - 9 x + 7 = 4 x^{2}

Move

4 x^{2}

to the left side

2 x^{2} - 9 x + 7 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 9 ) \pm ( - 9 ) ^{2} - 4 \cdot 2 \cdot 7}{2 \cdot 2}

(- 9)^{2} - 4 \cdot 2 \cdot 7 = 5

x_{1, 2} = \frac{- ( - 9 ) \pm 5}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- ( - 9 ) + 5}{2 \cdot 2}, x_{2} = \frac{- ( - 9 ) - 5}{2 \cdot 2}

x = \frac{- ( - 9 ) + 5}{2 \cdot 2} : \frac{7}{2}

x = \frac{- ( - 9 ) - 5}{2 \cdot 2} : 1

The solutions to the quadratic equation are:

x = \frac{7}{2}, x = 1

- 10 = 2 q

x^{2} + 3 x y

x^{2} - 2 x + 1 = 16

5 x + 7 \geq 2

x^{2} + x - 2 (x^{2} + 3 x) > 0

What are the solutions to the equation 6x^2-4x+7=4x^2+5x ?
The solutions to the equation 6x^2-4x+7=4x^2+5x are x= 7/2 ,x=1
Find the zeros of 6x^2-4x+7=4x^2+5x
The zeros of 6x^2-4x+7=4x^2+5x are x= 7/2 ,x=1