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# How would you solve this equation: 2x^2 + 6x -95 = 0? Topic: How would you solve this equation: 2x^2 + 6x -95 = 0?
July 18, 2019 / By Eamon
Question: I can remember two different ways to find the answer for this type of question but both are giving me different answers. How would you solve this equation? ## Best Answers: How would you solve this equation: 2x^2 + 6x -95 = 0? Bysshe | 10 days ago
A problem like this is rather hard to factor or even impossible, so I would use the quadratic formula in this case. Formula => x = (-b +/- [b^2 - 4*a*c]^(1/2) ) / (2*a) Specifically: x = (-6 +/- [6^2-4*2*-95]^(1/2) ) /(2*2) => x = 5.5534 or -8.5534 To pick either of those two roots as one specific answer for a certain problem, you would need the graph or other initial conditions.
👍 268 | 👎 10
Did you like the answer? How would you solve this equation: 2x^2 + 6x -95 = 0? Share with your friends Originally Answered: Graphing equation question? how would solve this linear equation?
2x + 3(11/3) = 12 3/1 * 11/3 = 33/3 When multiplying, just multiply across, numerator to numerator and denominator to denominator. (I know I put a 1 under the 3, all whole numbers have a 1 under them) (example : 3/1 = 3) or ( lets say 5 in fraction form is 5/1 ) (Another fraction example 2/5 * 3/4 mult across and you get 2 * 3 = 6 and 5 * 4 = 20 answer 6/20 then, simplify and get 3/10) Now simplify 33/3 = 11 2x + 11 = 12 Now subtract 11 from both sides. and you get 2x = 1 2x = 1 Now divide 2 on both sides and you get x = 1/2 ordered pair is (1/2 , 11/3) Originally Answered: Graphing equation question? how would solve this linear equation?
2x + 3(11/3) = 12 look at 3(11/3) = (3 * 11) / 3 = 11 ( 3/3) = 11 that gives you 2x + 11 = 12 2x = 1 x = 1/2 the point (1/2 , 11/3) is a solution to the equation remember, when you multiply fractions, you multiply the tops and multiply the bottoms if there isn't a bottom for one of the fractions, treat it like 1 3(11/3) = (3/1)(11/3) = (3*11) / (1 * 3) = 33 / 3 = 11 Aldwen
yes you should use the quadratic formula, which is: -b +/- the square root of b^2 - 4ac ________________________________ 2a plug in a=2, b=6, and c= -95. if you have a scientific calculator, solve on that. if not, there are some great online ones. Hope this helps! :)
👍 110 | 👎 9 Tallulah
a=2 b=6 c=-95 -6+/- sq rt 6^2-4(2)(-95)/4 -6+/- sq rt 36+ 760/ 4 -6+/- sq rt 796/ 4 -3+/- sq rt 796/ 2 -3+/- 2 sq rt 199/ 2 -3+2 sq rt 199/ 2 -3- 2 sq rt 199 / 2
👍 101 | 👎 8 Originally Answered: Solve this equation; like a quadratic equation.?
With substitution it becomes a quadratic. Let u = w² and rewrite as: u² + 7u - 18 = 0 (u+9)(u-2) = 0 u = -9 and 2 = w² w = ±3i and ±√2

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