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

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

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

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

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(x + 1)^{2} - 3 x = 3

x = 2, x = - 1

Solution steps

(x + 1)^{2} - 3 x = 3

Expand

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

x^{2} - x + 1 = 3

Move

3

to the left side

x^{2} - x - 2 = 0

Solve with the quadratic formula

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

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

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

x = 2, x = - 1

(x - 7) (x - 3) + 10 x = 30

(x - 2)^{2} - 64 = 0

(x + 5)^{2} - 3 (x + 5) - 4 = 0

(x^{2} + 5 x - 3) = (x + 2)

0 = t^{2} - t

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