What are the solutions to the equation (x+1)^2-5(x+1)+6=0 ?

The solutions to the equation (x+1)^2-5(x+1)+6=0 are x=2,x=1

Find the zeros of (x+1)^2-5(x+1)+6=0

The zeros of (x+1)^2-5(x+1)+6=0 are x=2,x=1

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(x + 1)^{2} - 5 (x + 1) + 6 = 0

x = 2, x = 1

Solution steps

(x + 1)^{2} - 5 (x + 1) + 6 = 0

Expand

(x + 1)^{2} - 5 (x + 1) + 6 : x^{2} - 3 x + 2

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

Solve with the quadratic formula

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

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

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

x = 2, x = 1

x^{2} + 8 x + 0 = 0

x, x^{2} + y^{2} + z^{2} + 4 x - 6 y + 2 z + 6 = 0

4 x^{2} + 3 = 323

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

q, q^{2} + 2 pq f^{2} - p^{2} = 0

What are the solutions to the equation (x+1)^2-5(x+1)+6=0 ?
The solutions to the equation (x+1)^2-5(x+1)+6=0 are x=2,x=1
Find the zeros of (x+1)^2-5(x+1)+6=0
The zeros of (x+1)^2-5(x+1)+6=0 are x=2,x=1