How do you solve #4x^2 + 5x =6#?
First, transpose 6 to the other side so the equation would end up in standard form (
Here, we can use the quadratic formula which is
We have to find out the variables
We substitute them to the formula and then do the math.
Solve until you stop here with this answer:
The square root of 121 is 11, so we end up with:
We then do the arithmetic,
The answers are
Here we end up with
If you want to check, substitute your solutions (the answers) to your formula:
The both solutions work because the equation is true (
Take note that all solutions might not work because they might end up undefined. If all solutions do not work, then there is no solution.
There you go! Hope this helps!
This can be completed by factoring.
Rearrange the formula to equal zero:
Multiply the first and last coefficients:
The possible factors of -24:
-24 , 1
-12 , 2
-8 , 3
-6 , 4
-4 , 6
-3 , 8
-2 , 12
-1 , 24
Add each of these numbers.
Only -3,8 adds to the middle coefficient, +5
The possible factors of the first coefficient are:
Since -3 is not divisible by the factor 2, the factors must be 4, 1
So the equation factors can be written as:
8 is divisible by 4, and -3 is divisible by 1, so
Check by FOIL multiplying: