What are the solutions to the equation x^2+(b+4)x=0 ?

The solutions to the equation x^2+(b+4)x=0 are x=0,x=-b-4

Find the zeros of x^2+(b+4)x=0

The zeros of x^2+(b+4)x=0 are x=0,x=-b-4

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x^{2} + (b + 4) x = 0

x = 0, x = - b - 4

Solution steps

x^{2} + (b + 4) x = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( b + 4 ) \pm ( b + 4 ) ^{2} - 4 \cdot 1 \cdot 0}{2 \cdot 1}

Simplify

(b + 4)^{2} - 4 \cdot 1 \cdot 0 : b + 4

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

Separate the solutions

x_{1} = \frac{- ( b + 4 ) + b + 4}{2 \cdot 1}, x_{2} = \frac{- ( b + 4 ) - ( b + 4 )}{2 \cdot 1}

x = \frac{- ( b + 4 ) + b + 4}{2 \cdot 1} : 0

x = \frac{- ( b + 4 ) - ( b + 4 )}{2 \cdot 1} : - b - 4

The solutions to the quadratic equation are:

x = 0, x = - b - 4

x^{2} + 4 x = 5

2 x^{2} + 7 = 31

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

(2 x - 1)^{2} - 225 = 0

3 x^{2} - 4 x = 3

What are the solutions to the equation x^2+(b+4)x=0 ?
The solutions to the equation x^2+(b+4)x=0 are x=0,x=-b-4
Find the zeros of x^2+(b+4)x=0
The zeros of x^2+(b+4)x=0 are x=0,x=-b-4