What are the solutions to the equation (x+1)^2-7(x+1)+10=0 ?

The solutions to the equation (x+1)^2-7(x+1)+10=0 are x=4,x=1

Find the zeros of (x+1)^2-7(x+1)+10=0

The zeros of (x+1)^2-7(x+1)+10=0 are x=4,x=1

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(x + 1)^{2} - 7 (x + 1) + 10 = 0

x = 4, x = 1

Solution steps

(x + 1)^{2} - 7 (x + 1) + 10 = 0

Expand

(x + 1)^{2} - 7 (x + 1) + 10 : x^{2} - 5 x + 4

x^{2} - 5 x + 4 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 5 ) \pm ( - 5 ) ^{2} - 4 \cdot 1 \cdot 4}{2 \cdot 1}

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

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

x = 4, x = 1

x^{2} + 35 = 12 x

x, \frac{2 x ^{2}}{3} = d g

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

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

- x^{2} + 3 x - 2

What are the solutions to the equation (x+1)^2-7(x+1)+10=0 ?
The solutions to the equation (x+1)^2-7(x+1)+10=0 are x=4,x=1
Find the zeros of (x+1)^2-7(x+1)+10=0
The zeros of (x+1)^2-7(x+1)+10=0 are x=4,x=1