anonymous
  • anonymous
Use the quadratic formula to solve the equation? 4x2 – 6x – 10 = 0 is anybody able to show me work?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
the quadratic formula is -b (±)Sqrt[b-4ac]/2a\[-b \pm \frac{ \sqrt{b-4ac}}{ 2a}\]
anonymous
  • anonymous
where ax^2 +bx +c just substitute and you will get your answer.
jim_thompson5910
  • jim_thompson5910
\[\Large x = \frac{-b\pm\sqrt{b^2-4ac}}{2a}\] \[\Large x = \frac{-(-6)\pm\sqrt{(-6)^2-4(4)(-10)}}{2(4)}\] \[\Large x = \frac{6\pm\sqrt{36-(-160)}}{8}\] I'll let you finish.

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anonymous
  • anonymous
I have all that work! I just get confused with the last part? like would i subtract 36-(-160)
anonymous
  • anonymous
36--160? well double negative is positive, so? 36+160
jim_thompson5910
  • jim_thompson5910
36-(-160) is the same as 36+160
anonymous
  • anonymous
so it would become 196?
jim_thompson5910
  • jim_thompson5910
it would
anonymous
  • anonymous
|dw:1352933217966:dw|
anonymous
  • anonymous
so would that be the answer?
jim_thompson5910
  • jim_thompson5910
you can keep going and there's a +- between the 6 and square root
jim_thompson5910
  • jim_thompson5910
you should have \[\Large x = \frac{6\pm\sqrt{196}}{8}\] but you can keep going
anonymous
  • anonymous
(6*14)/8 =(3*14)/4 =(3*7)/2 =21/2
anonymous
  • anonymous
yes i have that!
anonymous
  • anonymous
okay, what is wrong?
anonymous
  • anonymous
wait where did you get the 6*14/8
jim_thompson5910
  • jim_thompson5910
the square root of 196 is 14, so... \[\Large x = \frac{6\pm\sqrt{196}}{8}\] becomes \[\Large x = \frac{6\pm14}{8}\]
anonymous
  • anonymous
above is very clear and correct, so first you do the plus sign which is (6+14)/8 this is ONE OF YOUR x's the you do (6-14)/8 this is your other x value so your x values are: 5/2 and -1
anonymous
  • anonymous
THank you so much! I didnt know how to find that! but you just made it clear for me! thank you both for your help! (:
anonymous
  • anonymous
your most welcome,

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