What are the solutions to the equation (x-3)(x+1)+4=0 ?

The solutions to the equation (x-3)(x+1)+4=0 are x=1

Find the zeros of (x-3)(x+1)+4=0

The zeros of (x-3)(x+1)+4=0 are x=1

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(x - 3) (x + 1) + 4 = 0

x = 1

Solution steps

(x - 3) (x + 1) + 4 = 0

Expand

(x - 3) (x + 1) + 4 : x^{2} - 2 x + 1

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

Solve with the quadratic formula

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

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

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

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

\frac{- ( - 2 )}{2 \cdot 1} = 1

x = 1

The solution to the quadratic equation is:

x = 1

x^{2} + (- 1) x + (- 1) + 8 = 0

c = (2 x - 3) (2 x - 2) - 2 x (x - 5)

x^{2} - x - a = 0

(x + 1)^{2} = 4 x

- 6 x^{2} + 5 x - 1 = 0

What are the solutions to the equation (x-3)(x+1)+4=0 ?
The solutions to the equation (x-3)(x+1)+4=0 are x=1
Find the zeros of (x-3)(x+1)+4=0
The zeros of (x-3)(x+1)+4=0 are x=1