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=-13/3

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=-13/3

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(x + 2)^{2} + 3 (x + 1) (x + 4) = (x - 3) (x + 3) - 1

x = - 2, x = - \frac{13}{3}

Decimal Notation

x = - 2, x = - 4.33333 \dots

Solution steps

(x + 2)^{2} + 3 (x + 1) (x + 4) = (x - 3) (x + 3) - 1

Expand

(x + 2)^{2} + 3 (x + 1) (x + 4) : 4 x^{2} + 19 x + 16

Expand

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

4 x^{2} + 19 x + 16 = x^{2} - 10

Move

10

to the left side

4 x^{2} + 19 x + 26 = x^{2}

Move

x^{2}

to the left side

3 x^{2} + 19 x + 26 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 19 \pm 1 9 ^{2} - 4 \cdot 3 \cdot 26}{2 \cdot 3}

1 9^{2} - 4 \cdot 3 \cdot 26 = 7

x_{1, 2} = \frac{- 19 \pm 7}{2 \cdot 3}

Separate the solutions

x_{1} = \frac{- 19 + 7}{2 \cdot 3}, x_{2} = \frac{- 19 - 7}{2 \cdot 3}

x = \frac{- 19 + 7}{2 \cdot 3} : - 2

x = \frac{- 19 - 7}{2 \cdot 3} : - \frac{13}{3}

The solutions to the quadratic equation are:

x = - 2, x = - \frac{13}{3}

8 x^{2} - 22 x + 15 = 0

2 x - x^{2} = x^{2}

3 (5 x + 7)^{2} - 18 = 48

0.000006 x^{2} - 0.02 x + 25 = 85 - 0.025 x

3 x^{2} - 4 x - 16 = 0

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=-13/3
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=-13/3