.006x^2+.05x=.04?

Question

This problem doesn't seem solveable to me. I am using the theorem x = (-b±√(b^2-4ac))/2a.

Answer 1

Make it in easy form by multiply both sides by 1000, the equation becomes

6X² + 50 X = 40

6X² +50X - 40 =0

PUT a=6 b=50 and c=-40 in quadratic formula you mentioned in question, you will easily get the answer.

x = (-50 ± √ [50² -4(6)(-40)] ) / 2(6)

x= ( -50 ± 58.821)/12

x = 0.73 , x = -9.06

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you can also solve it directly by putting the decimal values (a=0.006, b=0.05, c= -0.04)in question but that will be a little difficult for calculation otherwise the answer will be definitely same and the formula definitely works for it. You might have done mistake in calculations.

Answer 2

Multiply both sides by 1000 and put in standard form to get
6x²+50x-40=0 Since b²-4ac = 36+4*6*40 = 996 there are two real roots and the quadratic formula works just fine.


Doug

Answer 3

It is solvable.

The answer may not be one with a known value.

Multiply both sides by 100 so that we may get n equation without decimals.

x^2 + 50 x - 40 = 0

a = 1 b = + 50 c = - 40

Applying the formula to find the roots,

one root is = [ -50 + {( 50^2 - 4* 1 * (-40)} ^ 1/2] Divided by 2

Another root is = [ -50 - { ( 50 * 2 - 4 * 1 * (-40) } ^ 1/2] divided by 2

Simplify & get the answers.

[-50 + (260) ^ 1/2 ] whole divided by 2
&

[-50 - (260) ^ 1/2] whole divided by 2

Answer 4

Remember, it should be set up in the form ax^2 + bx + c = 0.
So your formula should look like this simply when done.
.006x^2 + .05x + -.04 = 0
A=.006
B=.05
C= -.04
Soooo..
x= -(.05) ± √ [(.05)^2-4(.006)(-.04)] / 2(.006)
Anyways,
the roots are
0.735, -9.068

Answer 5

6x^2+50x-40=0
x1=[-50+(2500+960)]/12
x2=[-50-(2500+960)]/12

Answer 6

this problem has no roots