What are the solutions to the equation (x+1)^2+x(x-9)+2=0 ?

The solutions to the equation (x+1)^2+x(x-9)+2=0 are x=3,x= 1/2

Find the zeros of (x+1)^2+x(x-9)+2=0

The zeros of (x+1)^2+x(x-9)+2=0 are x=3,x= 1/2

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(x + 1)^{2} + x (x - 9) + 2 = 0

x = 3, x = \frac{1}{2}

Solution steps

(x + 1)^{2} + x (x - 9) + 2 = 0

Expand

(x + 1)^{2} + x (x - 9) + 2 : 2 x^{2} - 7 x + 3

2 x^{2} - 7 x + 3 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 7 ) \pm ( - 7 ) ^{2} - 4 \cdot 2 \cdot 3}{2 \cdot 2}

(- 7)^{2} - 4 \cdot 2 \cdot 3 = 5

x_{1, 2} = \frac{- ( - 7 ) \pm 5}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- ( - 7 ) + 5}{2 \cdot 2}, x_{2} = \frac{- ( - 7 ) - 5}{2 \cdot 2}

x = \frac{- ( - 7 ) + 5}{2 \cdot 2} : 3

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

The solutions to the quadratic equation are:

x = 3, x = \frac{1}{2}

6 x^{2} + 2 = 32

12 x^{2} + 50 = 0

3 x^{2} + x - 15 = 0

p^{2} + 8 p - 3 = 0

- 38 = - 4 c^{2} - 2

What are the solutions to the equation (x+1)^2+x(x-9)+2=0 ?
The solutions to the equation (x+1)^2+x(x-9)+2=0 are x=3,x= 1/2
Find the zeros of (x+1)^2+x(x-9)+2=0
The zeros of (x+1)^2+x(x-9)+2=0 are x=3,x= 1/2