Question

(11-2 с)(5с-6)=80-10(с-3)^2

Answer 1

$\displaystyle (11-2c)(5c-6)=80-10(c-3)^2\\55c-66-10c^2+12c=80-10(c^2-6c+9)\\55c-66-10c^2+12c=80-10c^2+60c-90\\55c-66+12c=80+60c-90\\67c-66=-10+60c\\67c-60c=-10+66\\7c=56\\c=8$

Answer 2

To solve the equation (11 - 2c)(5c - 6) = 80 - 10(c - 3)^2, follow these steps:

1. Expand both sides of the equation: (11 - 2c)(5c - 6) = 80 - 10(c - 3)^2

Expanding the left side using the distributive property: 55c - 66 - 10c^2 + 30c = 80 - 10(c^2 - 6c + 9)

2. Simplify both sides: Combine like terms on the left side: (55c - 66) - (10c^2 - 30c) = 80 - (10c^2 - 60c + 90)

55c - 66 - 10c^2 + 30c = 80 - 10c^2 + 60c - 90

3. Continue to simplify: Combine like terms on both sides of the equation: 55c - 66 - 10c^2 + 30c = -10c^2 + 60c - 10

4. Now, subtract 55c and add 66 to both sides to isolate the variable terms on one side: -10c^2 + 30c - 55c + 66 = -10c^2 + 60c - 10 - 55c + 66

-10c^2 - 25c + 66 = -10c^2 + 5c + 56

5. Notice that the -10c^2 terms cancel out on both sides of the equation: -25c + 66 = 5c + 56

6. Move all the c terms to one side of the equation and constants to the other side: -25c - 5c = 56 - 66

-30c = -10

7. Divide both sides by -30 to solve for c: c = (-10) / (-30)

c = 1/3

So, the solution to the equation (11 - 2c)(5c - 6) = 80 - 10(c - 3)^2 is c = 1/3.