What are the solutions to the equation 2x^2-5=3x^2-12x+31 ?

The solutions to the equation 2x^2-5=3x^2-12x+31 are x=6

Find the zeros of 2x^2-5=3x^2-12x+31

The zeros of 2x^2-5=3x^2-12x+31 are x=6

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

2 x^{2} - 5 = 3 x^{2} - 12 x + 31

x = 6

Solution steps

2 x^{2} - 5 = 3 x^{2} - 12 x + 31

Switch sides

3 x^{2} - 12 x + 31 = 2 x^{2} - 5

Move

5

to the left side

3 x^{2} - 12 x + 36 = 2 x^{2}

Move

2 x^{2}

to the left side

x^{2} - 12 x + 36 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 12 ) \pm ( - 12 ) ^{2} - 4 \cdot 1 \cdot 36}{2 \cdot 1}

(- 12)^{2} - 4 \cdot 1 \cdot 36 = 0

x_{1, 2} = \frac{- ( - 12 ) \pm 0}{2 \cdot 1}

x = \frac{- ( - 12 )}{2 \cdot 1}

\frac{- ( - 12 )}{2 \cdot 1} = 6

x = 6

The solution to the quadratic equation is:

x = 6

2 m^{2} = 50

(m - 1) x^{2} + 2 (m - 1) x - 1 = 0

p^{2} - 2 p + 1 = 0

x^{2} = - x + 6

x, 3 x^{2} - 2 x - 3 = 0

What are the solutions to the equation 2x^2-5=3x^2-12x+31 ?
The solutions to the equation 2x^2-5=3x^2-12x+31 are x=6
Find the zeros of 2x^2-5=3x^2-12x+31
The zeros of 2x^2-5=3x^2-12x+31 are x=6