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using the quadratic formula, what are the solutions to 6p2 + 11p + 5 = 0
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so a = 6 b = 11 c = 5 plug those values into the formula \[x = \frac{ -b + \sqrt{b^2 - 4ac} }{ 2a }\]
thank you, where did you get the 4 from? I am trying to understand this, I can do some of them but have problems with others :(
\[x = \frac{ -11 \pm \sqrt{11^2 - 4(6)(5)} }{ 2(6)}\] \[x = \frac{ -11\pm \sqrt{121 - 120} }{ 12 }\] \[x = \frac{ -11 \pm \sqrt{1} }{ 12 }\]
Okay, so in the quadratic formula, 4 is a number that is already there
thank you, I think I am getting the hang of it.........I hope :).........oh ok, I am having a brain freeze
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Ha, it happens. the actual formula is the first comment I put. So; \[x = \frac{ -b \pm \sqrt{b^2 - 4ac} }{ 2a}\]
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