What are the solutions to the equation 2x^2-2x-40=x^2+5x-10 ?

The solutions to the equation 2x^2-2x-40=x^2+5x-10 are x=10,x=-3

Find the zeros of 2x^2-2x-40=x^2+5x-10

The zeros of 2x^2-2x-40=x^2+5x-10 are x=10,x=-3

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

2 x^{2} - 2 x - 40 = x^{2} + 5 x - 10

x = 10, x = - 3

Solution steps

2 x^{2} - 2 x - 40 = x^{2} + 5 x - 10

Move

10

to the left side

2 x^{2} - 2 x - 30 = x^{2} + 5 x

Move

5 x

to the left side

2 x^{2} - 7 x - 30 = x^{2}

Move

x^{2}

to the left side

x^{2} - 7 x - 30 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 7 ) \pm ( - 7 ) ^{2} - 4 \cdot 1 \cdot ( - 30 )}{2 \cdot 1}

(- 7)^{2} - 4 \cdot 1 \cdot (- 30) = 13

x_{1, 2} = \frac{- ( - 7 ) \pm 13}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 7 ) + 13}{2 \cdot 1}, x_{2} = \frac{- ( - 7 ) - 13}{2 \cdot 1}

x = \frac{- ( - 7 ) + 13}{2 \cdot 1} : 10

x = \frac{- ( - 7 ) - 13}{2 \cdot 1} : - 3

The solutions to the quadratic equation are:

x = 10, x = - 3

(x^{2} + 3 x + 9) = 0

(q) = q^{2} + 50 q + 1000

p^{2} = 1

3 x^{2} + 5 x + 3 = 5

(2 x + 3)^{2} = 18

What are the solutions to the equation 2x^2-2x-40=x^2+5x-10 ?
The solutions to the equation 2x^2-2x-40=x^2+5x-10 are x=10,x=-3
Find the zeros of 2x^2-2x-40=x^2+5x-10
The zeros of 2x^2-2x-40=x^2+5x-10 are x=10,x=-3