What are the solutions to the equation (10-x)(-12-x)+40=0 ?

The solutions to the equation (10-x)(-12-x)+40=0 are x=8,x=-10

Find the zeros of (10-x)(-12-x)+40=0

The zeros of (10-x)(-12-x)+40=0 are x=8,x=-10

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(10 - x) (- 12 - x) + 40 = 0

x = 8, x = - 10

Solution steps

(10 - x) (- 12 - x) + 40 = 0

Expand

(10 - x) (- 12 - x) + 40 : x^{2} + 2 x - 80

x^{2} + 2 x - 80 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 2 \pm 2 ^{2} - 4 \cdot 1 \cdot ( - 80 )}{2 \cdot 1}

2^{2} - 4 \cdot 1 \cdot (- 80) = 18

x_{1, 2} = \frac{- 2 \pm 18}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 2 + 18}{2 \cdot 1}, x_{2} = \frac{- 2 - 18}{2 \cdot 1}

x = \frac{- 2 + 18}{2 \cdot 1} : 8

x = \frac{- 2 - 18}{2 \cdot 1} : - 10

The solutions to the quadratic equation are:

x = 8, x = - 10

x (4 x - 1) = 3 x (x - 2) + 6

x, x^{2} - k x + 16 = 0

x^{2} = - 1 - 4 x

3 x^{2} - 15 = 4 x

2 x^{2} - 10 x = - 7 x - 2

What are the solutions to the equation (10-x)(-12-x)+40=0 ?
The solutions to the equation (10-x)(-12-x)+40=0 are x=8,x=-10
Find the zeros of (10-x)(-12-x)+40=0
The zeros of (10-x)(-12-x)+40=0 are x=8,x=-10