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Sagot :
I will use the quadratic equation to solve this:
x= -b +(or minus, both) √b² + 4ac
--------------------------------------------
2a
So, your numbers would be: a = 3
b = 13
c = 10
Then, solve: x = -13 +(or minus, both) √169 + 120
--------------------------------------------------
6
= -13 +(or minus, both) 17
------------------------------------
6
So, first do the plus: -13 + 17
--------------
6
= 2/3
Then do the minus: -13-17
------------
6
= 5
Your answer is: 2/3 and 5
( These problems always have two answers)
x= -b +(or minus, both) √b² + 4ac
--------------------------------------------
2a
So, your numbers would be: a = 3
b = 13
c = 10
Then, solve: x = -13 +(or minus, both) √169 + 120
--------------------------------------------------
6
= -13 +(or minus, both) 17
------------------------------------
6
So, first do the plus: -13 + 17
--------------
6
= 2/3
Then do the minus: -13-17
------------
6
= 5
Your answer is: 2/3 and 5
( These problems always have two answers)
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