What are the solutions to the equation x^2+4x-5=x-1 ?

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

Find the zeros of x^2+4x-5=x-1

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

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x^{2} + 4 x - 5 = x - 1

x = 1, x = - 4

Solution steps

x^{2} + 4 x - 5 = x - 1

Move

1

to the left side

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

Move

x

to the left side

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

Solve with the quadratic formula

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

3^{2} - 4 \cdot 1 \cdot (- 4) = 5

x_{1, 2} = \frac{- 3 \pm 5}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 3 + 5}{2 \cdot 1}, x_{2} = \frac{- 3 - 5}{2 \cdot 1}

x = \frac{- 3 + 5}{2 \cdot 1} : 1

x = \frac{- 3 - 5}{2 \cdot 1} : - 4

The solutions to the quadratic equation are:

x = 1, x = - 4

x^{2} - 5 x = - 3

2 (5 x - 1)^{2} + 28 = 30

2 x (1 + \frac{1}{2} x) = x (\frac{1}{2} x + 1) + 12

x^{2} - 1.3 x - 0.3 = 0

32 - 6 x^{2} = 0

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