Questions on Algebra: Quadratic Equation answered by real tutors!

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Question 150810This question is from textbook Blitzer (college Algebra
: How do I break down the problem and solve it, 3x^2+19x-14=0 using the Quadratic Equation?This question is from textbook Blitzer (college Algebra
: How do I break down the problem and solve it, 3x^2+19x-14=0 using the Quadratic Equation?
Answer by jim_thompson5910(8066) About Me  (Show Source):
You can put this solution on YOUR website!
From 3x^2+19x-14, we can see that a=3, b=19, and c=-14


x = (-b +- sqrt( b^2-4ac ))/(2a) Start with the quadratic formula


x = (-(19) +- sqrt( (19)^2-4(3)(-14) ))/(2(3)) Plug in a=3, b=19, and c=-14


x = (-19 +- sqrt( 361-4(3)(-14) ))/(2(3)) Square 19 to get 361.


x = (-19 +- sqrt( 361--168 ))/(2(3)) Multiply 4(3)(-14) to get -168


x = (-19 +- sqrt( 361+168 ))/(2(3)) Rewrite sqrt(361--168) as sqrt(361+168)


x = (-19 +- sqrt( 529 ))/(2(3)) Add 361 to 168 to get 529


x = (-19 +- sqrt( 529 ))/(6) Multiply 2 and 3 to get 6.


x = (-19 +- 23)/(6) Take the square root of 529 to get 23.


x = (-19 + 23)/(6) or x = (-19 - 23)/(6) Break up the expression.


x = (4)/(6) or x = (-42)/(6) Combine like terms.


x = 2/3 or x = -7 Simplify.


So our answers are x = 2/3 or x = -7