What are the solutions to the equation (x+3)^2+(x+2)(x-2)=x(x-1)-5 ?

The solutions to the equation (x+3)^2+(x+2)(x-2)=x(x-1)-5 are x=-2,x=-5

Find the zeros of (x+3)^2+(x+2)(x-2)=x(x-1)-5

The zeros of (x+3)^2+(x+2)(x-2)=x(x-1)-5 are x=-2,x=-5

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(x + 3)^{2} + (x + 2) (x - 2) = x (x - 1) - 5

x = - 2, x = - 5

Solution steps

(x + 3)^{2} + (x + 2) (x - 2) = x (x - 1) - 5

Expand

(x + 3)^{2} + (x + 2) (x - 2) : 2 x^{2} + 6 x + 5

Expand

x (x - 1) - 5 : x^{2} - x - 5

2 x^{2} + 6 x + 5 = x^{2} - x - 5

Move

5

to the left side

2 x^{2} + 6 x + 10 = x^{2} - x

Move

x

to the left side

2 x^{2} + 7 x + 10 = x^{2}

Move

x^{2}

to the left side

x^{2} + 7 x + 10 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 7 \pm 7 ^{2} - 4 \cdot 1 \cdot 10}{2 \cdot 1}

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

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

x = - 2, x = - 5

u^{2} - 25 u + 144 = 0

x^{2} = 4 x + 32

x^{2} - 1 = 0

x, f = x^{2} - h^{2}

3 x^{2} = 13 x - 10

What are the solutions to the equation (x+3)^2+(x+2)(x-2)=x(x-1)-5 ?
The solutions to the equation (x+3)^2+(x+2)(x-2)=x(x-1)-5 are x=-2,x=-5
Find the zeros of (x+3)^2+(x+2)(x-2)=x(x-1)-5
The zeros of (x+3)^2+(x+2)(x-2)=x(x-1)-5 are x=-2,x=-5