What are the solutions to the equation (x+2)2+3(x+1)(x+4)=(x-3)(x+3)-1 ?

The solutions to the equation (x+2)2+3(x+1)(x+4)=(x-3)(x+3)-1 are x=-2,x=-13/2

Find the zeros of (x+2)2+3(x+1)(x+4)=(x-3)(x+3)-1

The zeros of (x+2)2+3(x+1)(x+4)=(x-3)(x+3)-1 are x=-2,x=-13/2

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(x + 2) 2 + 3 (x + 1) (x + 4) = (x - 3) (x + 3) - 1

x = - 2, x = - \frac{13}{2}

Decimal Notation

x = - 2, x = - 6.5

Solution steps

(x + 2) \cdot 2 + 3 (x + 1) (x + 4) = (x - 3) (x + 3) - 1

Expand

(x + 2) \cdot 2 + 3 (x + 1) (x + 4) : 3 x^{2} + 17 x + 16

Expand

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

3 x^{2} + 17 x + 16 = x^{2} - 10

Move

10

to the left side

3 x^{2} + 17 x + 26 = x^{2}

Move

x^{2}

to the left side

2 x^{2} + 17 x + 26 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 17 \pm 1 7 ^{2} - 4 \cdot 2 \cdot 26}{2 \cdot 2}

1 7^{2} - 4 \cdot 2 \cdot 26 = 9

x_{1, 2} = \frac{- 17 \pm 9}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- 17 + 9}{2 \cdot 2}, x_{2} = \frac{- 17 - 9}{2 \cdot 2}

x = \frac{- 17 + 9}{2 \cdot 2} : - 2

x = \frac{- 17 - 9}{2 \cdot 2} : - \frac{13}{2}

The solutions to the quadratic equation are:

x = - 2, x = - \frac{13}{2}

2 (x + 5)^{2} = 8

9 x^{2} - 16 x - 7 = 2 x + 9

72 x^{2} - 48 x = 0

4 x^{2} + 2 = 10

\frac{x ^{2}}{4} = x

What are the solutions to the equation (x+2)2+3(x+1)(x+4)=(x-3)(x+3)-1 ?
The solutions to the equation (x+2)2+3(x+1)(x+4)=(x-3)(x+3)-1 are x=-2,x=-13/2
Find the zeros of (x+2)2+3(x+1)(x+4)=(x-3)(x+3)-1
The zeros of (x+2)2+3(x+1)(x+4)=(x-3)(x+3)-1 are x=-2,x=-13/2