What are the solutions to the equation (x+2)^2+11(x+2)-12=0 ?

The solutions to the equation (x+2)^2+11(x+2)-12=0 are x=-1,x=-14

Find the zeros of (x+2)^2+11(x+2)-12=0

The zeros of (x+2)^2+11(x+2)-12=0 are x=-1,x=-14

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(x + 2)^{2} + 11 (x + 2) - 12 = 0

x = - 1, x = - 14

Solution steps

(x + 2)^{2} + 11 (x + 2) - 12 = 0

Expand

(x + 2)^{2} + 11 (x + 2) - 12 : x^{2} + 15 x + 14

x^{2} + 15 x + 14 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 15 \pm 1 5 ^{2} - 4 \cdot 1 \cdot 14}{2 \cdot 1}

1 5^{2} - 4 \cdot 1 \cdot 14 = 13

x_{1, 2} = \frac{- 15 \pm 13}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 15 + 13}{2 \cdot 1}, x_{2} = \frac{- 15 - 13}{2 \cdot 1}

x = \frac{- 15 + 13}{2 \cdot 1} : - 1

x = \frac{- 15 - 13}{2 \cdot 1} : - 14

The solutions to the quadratic equation are:

x = - 1, x = - 14

y, y^{2} - 2 x y + 9 = 0

3 x^{2} - 2 x = 2 x + 7

x^{2} - 5 x - 15 = 0

a, (a + b) (a - b) = b (a - b)

A (y) = 240 y - 2 y^{2}

What are the solutions to the equation (x+2)^2+11(x+2)-12=0 ?
The solutions to the equation (x+2)^2+11(x+2)-12=0 are x=-1,x=-14
Find the zeros of (x+2)^2+11(x+2)-12=0
The zeros of (x+2)^2+11(x+2)-12=0 are x=-1,x=-14