What are the solutions to the equation (x+3)^2+7(x+3)-18=0 ?

The solutions to the equation (x+3)^2+7(x+3)-18=0 are x=-1,x=-12

Find the zeros of (x+3)^2+7(x+3)-18=0

The zeros of (x+3)^2+7(x+3)-18=0 are x=-1,x=-12

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(x + 3)^{2} + 7 (x + 3) - 18 = 0

x = - 1, x = - 12

Solution steps

(x + 3)^{2} + 7 (x + 3) - 18 = 0

Expand

(x + 3)^{2} + 7 (x + 3) - 18 : x^{2} + 13 x + 12

x^{2} + 13 x + 12 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 13 \pm 1 3 ^{2} - 4 \cdot 1 \cdot 12}{2 \cdot 1}

1 3^{2} - 4 \cdot 1 \cdot 12 = 11

x_{1, 2} = \frac{- 13 \pm 11}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 13 + 11}{2 \cdot 1}, x_{2} = \frac{- 13 - 11}{2 \cdot 1}

x = \frac{- 13 + 11}{2 \cdot 1} : - 1

x = \frac{- 13 - 11}{2 \cdot 1} : - 12

The solutions to the quadratic equation are:

x = - 1, x = - 12

2^{\frac{8}{3}}

- 4 = 5 + x

- 2 g + 4 \leq - 6 g - 9

x^{4} - 20 x^{2} + 64 \geq 0

(\frac{6}{5})^{5 (- \frac{1}{2})}

What are the solutions to the equation (x+3)^2+7(x+3)-18=0 ?
The solutions to the equation (x+3)^2+7(x+3)-18=0 are x=-1,x=-12
Find the zeros of (x+3)^2+7(x+3)-18=0
The zeros of (x+3)^2+7(x+3)-18=0 are x=-1,x=-12