What are the solutions to the equation 2(x-1)(x+3)=(x+5)(x-5) ?

The solutions to the equation 2(x-1)(x+3)=(x+5)(x-5) are x=-2+sqrt(15)i,x=-2-sqrt(15)i

Find the zeros of 2(x-1)(x+3)=(x+5)(x-5)

The zeros of 2(x-1)(x+3)=(x+5)(x-5) are x=-2+sqrt(15)i,x=-2-sqrt(15)i

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

2 (x - 1) (x + 3) = (x + 5) (x - 5)

x = - 2 + 15 i, x = - 2 - 15 i

Solution steps

2 (x - 1) (x + 3) = (x + 5) (x - 5)

Expand

2 (x - 1) (x + 3) : 2 x^{2} + 4 x - 6

Expand

(x + 5) (x - 5) : x^{2} - 25

2 x^{2} + 4 x - 6 = x^{2} - 25

Move

25

to the left side

2 x^{2} + 4 x + 19 = x^{2}

Move

x^{2}

to the left side

x^{2} + 4 x + 19 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 4 \pm 4 ^{2} - 4 \cdot 1 \cdot 19}{2 \cdot 1}

Simplify

4^{2} - 4 \cdot 1 \cdot 19 : 215 i

x_{1, 2} = \frac{- 4 \pm 2 15 i}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 4 + 2 15 i}{2 \cdot 1}, x_{2} = \frac{- 4 - 2 15 i}{2 \cdot 1}

x = \frac{- 4 + 2 15 i}{2 \cdot 1} : - 2 + 15 i

x = \frac{- 4 - 2 15 i}{2 \cdot 1} : - 2 - 15 i

The solutions to the quadratic equation are:

x = - 2 + 15 i, x = - 2 - 15 i

x^{2} - x + 3 = 4 x^{2} - 5

- x^{2} + 3 x - 2

[f (a)]^{2} = 3 x^{2} - x + 2

- 4 x^{2} + 9 x - 3 = 0

- 0.5 x^{2} + 3.2 x + 22 = 24

What are the solutions to the equation 2(x-1)(x+3)=(x+5)(x-5) ?
The solutions to the equation 2(x-1)(x+3)=(x+5)(x-5) are x=-2+sqrt(15)i,x=-2-sqrt(15)i
Find the zeros of 2(x-1)(x+3)=(x+5)(x-5)
The zeros of 2(x-1)(x+3)=(x+5)(x-5) are x=-2+sqrt(15)i,x=-2-sqrt(15)i