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

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

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

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

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

x = 1

Solution steps

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

Expand

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

(- 3 y)^{2} + y^{2} = 40

4 x + x \cdot 4 (x - 2) = (3 x + 15) (x - 2)

16 t^{2} - 64 t + 48 = 0

2 x^{2} - 3 x = - 6

a, u^{3} = a^{2} t

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