What are the solutions to the equation (x+1)^2+(x+1)-12=0 ?

The solutions to the equation (x+1)^2+(x+1)-12=0 are x=2,x=-5

Find the zeros of (x+1)^2+(x+1)-12=0

The zeros of (x+1)^2+(x+1)-12=0 are x=2,x=-5

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(x + 1)^{2} + (x + 1) - 12 = 0

x = 2, x = - 5

Solution steps

(x + 1)^{2} + (x + 1) - 12 = 0

Expand

(x + 1)^{2} + (x + 1) - 12 : x^{2} + 3 x - 10

x^{2} + 3 x - 10 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 3 \pm 3 ^{2} - 4 \cdot 1 \cdot ( - 10 )}{2 \cdot 1}

3^{2} - 4 \cdot 1 \cdot (- 10) = 7

x_{1, 2} = \frac{- 3 \pm 7}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 3 + 7}{2 \cdot 1}, x_{2} = \frac{- 3 - 7}{2 \cdot 1}

x = \frac{- 3 + 7}{2 \cdot 1} : 2

x = \frac{- 3 - 7}{2 \cdot 1} : - 5

The solutions to the quadratic equation are:

x = 2, x = - 5

- 3 x^{2} + 21 x - 36 = 0

α, x^{2} + α^{2} y^{2} = 4 α^{2}

5 w^{2} = 30 w

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

9 x^{2} - 35 = 14

What are the solutions to the equation (x+1)^2+(x+1)-12=0 ?
The solutions to the equation (x+1)^2+(x+1)-12=0 are x=2,x=-5
Find the zeros of (x+1)^2+(x+1)-12=0
The zeros of (x+1)^2+(x+1)-12=0 are x=2,x=-5