What are the solutions to the equation (3-x)(-1-x)-5=0 ?

The solutions to the equation (3-x)(-1-x)-5=0 are x=4,x=-2

Find the zeros of (3-x)(-1-x)-5=0

The zeros of (3-x)(-1-x)-5=0 are x=4,x=-2

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

x = 4, x = - 2

Solution steps

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

Expand

(3 - x) (- 1 - x) - 5 : x^{2} - 2 x - 8

x^{2} - 2 x - 8 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 2 ) \pm ( - 2 ) ^{2} - 4 \cdot 1 \cdot ( - 8 )}{2 \cdot 1}

(- 2)^{2} - 4 \cdot 1 \cdot (- 8) = 6

x_{1, 2} = \frac{- ( - 2 ) \pm 6}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 2 ) + 6}{2 \cdot 1}, x_{2} = \frac{- ( - 2 ) - 6}{2 \cdot 1}

x = \frac{- ( - 2 ) + 6}{2 \cdot 1} : 4

x = \frac{- ( - 2 ) - 6}{2 \cdot 1} : - 2

The solutions to the quadratic equation are:

x = 4, x = - 2

x^{2} + 25 x + 150 = 0

(4 x + 5)^{2} = 5

(m + 3) (m - 3) = 0

207 = 2 x^{2} + 7

x, f^{- 1} (x) = x^{2} + 4 x

What are the solutions to the equation (3-x)(-1-x)-5=0 ?
The solutions to the equation (3-x)(-1-x)-5=0 are x=4,x=-2
Find the zeros of (3-x)(-1-x)-5=0
The zeros of (3-x)(-1-x)-5=0 are x=4,x=-2