What are the solutions to the equation (x-5)(x+1)+5=0 ?

The solutions to the equation (x-5)(x+1)+5=0 are x=4,x=0

Find the zeros of (x-5)(x+1)+5=0

The zeros of (x-5)(x+1)+5=0 are x=4,x=0

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(x - 5) (x + 1) + 5 = 0

x = 4, x = 0

Solution steps

(x - 5) (x + 1) + 5 = 0

Expand

(x - 5) (x + 1) + 5 : x^{2} - 4 x

x^{2} - 4 x = 0

Solve with the quadratic formula

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

(- 4)^{2} - 4 \cdot 1 \cdot 0 = 4

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

x = 4, x = 0

x^{2} = 1 0^{2} + 1 0^{2}

7 = - x^{2} - 9 x

x^{2} - \frac{3}{4} x - \frac{27}{64} = 0

x, 9 x^{2} + y^{2} + 4 z^{2} - 18 x + 2 y + 16 z = 10

10 x^{2} = - 19 x + 15

What are the solutions to the equation (x-5)(x+1)+5=0 ?
The solutions to the equation (x-5)(x+1)+5=0 are x=4,x=0
Find the zeros of (x-5)(x+1)+5=0
The zeros of (x-5)(x+1)+5=0 are x=4,x=0