What are the solutions to the equation solvefor x,x^2+2x-1=2 ?

The solutions to the equation solvefor x,x^2+2x-1=2 are x=1,x=-3

Find the zeros of solvefor x,x^2+2x-1=2

The zeros of solvefor x,x^2+2x-1=2 are x=1,x=-3

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

x = 1, x = - 3

Solution steps

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

Move

2

to the left side

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

Solve with the quadratic formula

x_{1, 2} = \frac{- 2 \pm 2 ^{2} - 4 \cdot 1 \cdot ( - 3 )}{2 \cdot 1}

2^{2} - 4 \cdot 1 \cdot (- 3) = 4

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

Separate the solutions

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

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

x = \frac{- 2 - 4}{2 \cdot 1} : - 3

The solutions to the quadratic equation are:

x = 1, x = - 3

x^{2} = \frac{- b \pm b ^{2} - 4 a c}{2 a}

(3 w - 5) w = 28

t^{2} + 2 t + 6 = 7 t + 2

x^{2} - 16 x - 36 = 0

k^{2} - 2 k = - 2

What are the solutions to the equation solvefor x,x^2+2x-1=2 ?
The solutions to the equation solvefor x,x^2+2x-1=2 are x=1,x=-3
Find the zeros of solvefor x,x^2+2x-1=2
The zeros of solvefor x,x^2+2x-1=2 are x=1,x=-3