Algebra help please!?

Here's my equation; 3a^2 -13a=10 I understand the next step, -10 from both sides to get to here;

3a^2-13a-10=0. Now the mml software has a "Help me Solve this" button and when I click on that it shows me this; 3a^2-13a-10=0


I don't understand where the +2 comes from or know what to do in order to fill in the ?.

Can someone please explain what I need to be doing here, I know I have to factor the 10 right, so my solution set should either be {2, 5} or {1, 10} or am I just way off here? Please help!

4 Answers

  • 9 years ago
    Favorite Answer

    First you it to factor it...

    3a^2-13a-10=0 <----- I prefer factoring using decomposition, so thats how I'll show you how to do it.

    First step is you want to find two numbers which same a sum of -13(middle term) and a product of

    -30(you multiply the first and last term together).

    The two numbers i found were -15 and +2.

    Now you replace the -13 with these two numbers, so you get,

    3a^2 - 15a +2a -10 = 0 <----- now you need to factor the first two terms and the last two terms. I like

    -------------- -----------

    to under line then so I know what which numbers i'm factoring.

    When you factor them you will get,

    3a(a-5) +2(a-5) <----- the numbers in both set of brackets should be the same.

    Now you take the two numbers following each set of brackets and put them together and they make one factor, and the other factor would be the numbers already in the brackets.

    (3a+2) (a-5) <------ thats where the +2 came from!

    a= -2/3 a= 5

    It's hard for me to explain this, but I hope this helps!

  • Luis
    Lv 6
    9 years ago

    Dear Jim,


    Move 10 to the left-hand side of the equation by subtracting it from both sides. The goal is to have all terms on the left-hand side equal to 0.


    Use the quadratic formula to find the solutions. In this case, the values are a=3, b=-13, and c=-10.

    a=(-b\~(b^(2)-4ac))/(2a) where aa^(2)+ba+c=0

    Substitute in the values of a=3, b=-13, and c=-10.


    Multiply -1 by the -13 inside the parentheses.


    Squaring an expression is the same as multiplying the expression by itself 2 times.


    Multiply -13 by -13 to get 169.


    Remove the parentheses around the expression 169.


    Multiply 3 by -10 to get -30.


    Multiply -4 by the -30 inside the parentheses.


    Multiply 4 by 30 to get 120.


    Remove the parentheses around the expression 169.


    Add 120 to 169 to get 289.


    Pull all perfect square roots out from under the radical. In this case, remove the 17 because it is a perfect square.


    Multiply 2 by the 3 inside the parentheses.


    Multiply 2 by 3 to get 6.


    First, solve the + portion of \.


    Add 17 to 13 to get 30.


    Cancel the common factor of 6 in (30)/(6).


    Remove the common factors that were cancelled out.


    Next, solve the - portion of \.


    Subtract 17 from 13 to get -4.


    Move the minus sign from the numerator to the front of the expression.


    Cancel the common factor of 2 in -(4)/(6) since -(4)/(6)=((-2*2))/((3*2)).


    Remove the common factors that were cancelled out.


    The final answer is the combination of both solutions.


    Source(s): Algebra Solved!
  • Anonymous
    9 years ago

    i'm confused what the ^ is. do you mean a to the second power? or is it a different symbol???

    if it is 3a to the second power. well, i don't know how to show the work, but the answer is a = 5

    if you do a to the second power, ofr 5x5 is 25 x 3 is 75 minus 13x5 or 65 = 10

    i only got the solution from trying each number 1, 2, 3, 4, 5. and bingo 5 worked, but i don't know how to show how to come to that conclusion. i forgot. sorry.

  • 9 years ago

    well for this one you need the formula

    x = -b± (b²-4ac)



    but its the root of the whole equation, the root of the (b²-4ac) over 2a

    so youd get

    13±(13² -4 (3) (-10)



    = 13± [root] (169 + 120)



    =13 + 6.9 = 19.9

    =13 - 6.9 = 6.1

    sorry, you were way off, but hopefully this should help:)x

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