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

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

Find the zeros of x^2+(n+3)x+n+2=0

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

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x^{2} + (n + 3) x + n + 2 = 0

x = - 1, x = - n - 2

Solution steps

x^{2} + (n + 3) x + n + 2 = 0

Solve with the quadratic formula

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

Simplify

(n + 3)^{2} - 4 \cdot 1 \cdot (n + 2) : n + 1

x_{1, 2} = \frac{- ( n + 3 ) \pm ( n + 1 )}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( n + 3 ) + n + 1}{2 \cdot 1}, x_{2} = \frac{- ( n + 3 ) - ( n + 1 )}{2 \cdot 1}

x = \frac{- ( n + 3 ) + n + 1}{2 \cdot 1} : - 1

x = \frac{- ( n + 3 ) - ( n + 1 )}{2 \cdot 1} : - n - 2

The solutions to the quadratic equation are:

x = - 1, x = - n - 2

f, f = f (x + 2) - 3

x^{2} + 9 = 8 x

x, x^{2} y + y^{2} x = 6

p, x^{2} p^{2} + x y p - 6 y^{2} = 0

x^{2} - 9 x + 16 = 0

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