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

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

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

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

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(x + 1) (x - 3) = - 3

x = 2, x = 0

Solution steps

(x + 1) (x - 3) = - 3

Expand

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

x^{2} - 2 x - 3 = - 3

Move

3

to the left side

x^{2} - 2 x = 0

Solve with the quadratic formula

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

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

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

x = 2, x = 0

c, a = b + c^{2} d

x, x^{2} + 4 y^{2} = 0

n^{2} + 6 n - 16 = 0

x^{2} + \frac{3}{4} x - 1

y^{2} = - 4 y

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