SOLUTION: Question: Solve 4x^2+3x=10 (a) (2,-4/5) (b) (-2, 4/5) (c) (2, -5/4) (d) (-2, 5/4) Please choose one the answers from above- thanks in advance

Question 30239: Question:
Solve 4x^2+3x=10
(a) (2,-4/5)
(b) (-2, 4/5)
(c) (2, -5/4)
(d) (-2, 5/4)
Please choose one the answers from above- thanks in advance
Answer by checkley71(8403) (Show Source): You can put this solution on YOUR website!
4x~2+3x=10 or 4x~2+3x-10=0 Factors 4 are 1,2,&4. Factors of 10 are 1,2,5&10
because the 10is negative -- one of these factors must be a negative number and the other on a positive number. factors of 4 must be both positive or both negative.
Therefore the only combination to get a middle term of +3 are (4x-5)(x+2)=0.
Or 4x-5=0 or 4x=5 or x=5/4 and x+2=0 or x=-2.
(d) (-2, 5/4)
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