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JakeV8 Group TitleBest ResponseYou've already chosen the best response.1
Start by getting all the number terms together... subtract 125 from both sides so you have everything on the right, and only "0" on the left: 125 = 0.01x^2 + 0.05x + 107 125  125 = 0.01x^2 + 0.05x + 107  125 0 = 0.01x^2 + 0.05x 18
 one year ago

kenneyfamily Group TitleBest ResponseYou've already chosen the best response.0
ok and then what?
 one year ago

JakeV8 Group TitleBest ResponseYou've already chosen the best response.1
You might like the way it looks better if you multiply the whole thing through by 100... it would give you whole numbers, at least: 0 = 0.01x^2 + 0.05x  18 Multiply by 100: 0 = x^2 + 5x  1800 At this point, unless you can see how to factor it, you probably need to plug it into the quadratic formula to get the solutions for x.
 one year ago

JakeV8 Group TitleBest ResponseYou've already chosen the best response.1
Quadratic formula says, if you have 0 = ax^2 + bx + c, then the coefficients of the terms are "a", "b", and "c", In this problem, it's: 0 = x^2 + 5x  1800 so: a = 1, b = 5, and c = 1800.
 one year ago

JakeV8 Group TitleBest ResponseYou've already chosen the best response.1
Then the two solutions for x are given by plugging those a, b, and c values into the quadratic formula: \[\frac{ b \pm \sqrt{b ^{2}4ac} }{ 2a }\]
 one year ago

JakeV8 Group TitleBest ResponseYou've already chosen the best response.1
\[\frac{ b \pm \sqrt{b^{2} 4ac} }{ 2a } = \frac{ 5 \pm \sqrt{5^{2}  4(1)(1800)} }{ 2(1) } = \frac{ 5 \pm \sqrt{25 + 7200} }{ 2 }\]
 one year ago

JakeV8 Group TitleBest ResponseYou've already chosen the best response.1
Under the square root, you end up with 7225, and the square root of 7225 is 85. So you have two solutions for x: (5 + 85)/2 and (5  85)/2
 one year ago
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