WILL GIVE A MEDAL PLEASE HELP.
Give the values of a, b, and c needed to write the equation's general form.

2/3(X-4)(X+5)=1

A. A = 2/3; B = 1; C = -20

B. A = 2; B = 2; C = 43

C. A = 2; B = 2; C = -43

Question

WILL GIVE A MEDAL PLEASE HELP.
Give the values of a, b, and c needed to write the equation's general form.

2/3(X-4)(X+5)=1



A. A = 2/3; B = 1; C = -20

B. A = 2; B = 2; C = 43

C. A = 2; B = 2; C = -43

anonymous · Answer

a, b and c ? O.o its nowhere in the equation!

anonymous · Answer

Okay see, as per I understand find the value of x right? 
take the denominator of the LHS to the RHS. The value of the denominator remains unchanged cause we've just multiplied it by 1

anonymous · Answer

Okay so hereon we get a quadratic equation. Remember to get a,b and c Either Rhs or LHS should be zero. So next, take 2 (which is in the LHS) to the RHS and simplify. Now compare the coefficients of x^2 , x and constant to get the values of a, b , c respectively :-)