How do I solve x^2 + 0.5x - 0.1 = 0 for x?

x^2 + 0.5x - 0.1 = 0
10(x^2 + 0.5x - 0.1) = 10(0)
10x^2 + 5x - 1 = 0
x = [-b ±√(b^2 - 4ac)]/2a

a = 10
b = 5
c = -1

x = [-5 ±√(25 + 40]/20
x = [-5 ±√65]/20
x = [-5 ±8.06]/20 (approx.)

x = [-5 + 8.06]/20
x = 3.06/20
x = 0.153

x = [-5 - 8.06]/20
x = -13.06/20
x = -0.653

∴ x = 0.153 , -0.653 (approx.)

x^2 + 0.5x - 0.1 = 0
A = 1, B = 0.5, C = -0.1
using quadratic formula;
-B +/- sqrt B^2 - 4AC all over 2A...
= -0.5 +/- sqrt [(0.5)^2 - 4(1)(-0.1)] all over 2(1)
= -0.5 +/- sqrt (0.25 + 0.4) all over 2
= -0.5 +/- sqrt (0.65) all over 2
= (-0.5 + 0.81)/2
= 0.31/2
= 0.155
...OR...
= (-0.5 - 0.81)/2
= -1.31/2
= -0.655

Use the quadratic formula:

x = (-b +/- sqrt(b^2 - 4ac))/2a

Here,

x = (-0.5 +/- sqrt(0.5^2 - 4(1)(-0.1)))/2(1)
= (-0.5 +/- sqrt(0.25 + 0.4))/2
= (-0.5 +/- sqrt(0.65))/2
= -0.25 +/- sqrt(0.65)/2

So, x is either
-0.25 + sqrt(0.65)/2
or
-0.25 - sqrt(0.65)/2

Enjoy!

Rob

x^2 + 0.5x - 0.1 = 0

x=[-0.5+/-√0.5^2+4*0.1]/2=
(-0.5+/-√0.65)/2=
(-0.5+/-0.8)/2=
x=0.15
x= -0.65.

-0.5+-sqrt(0.9)/2

10x² + 5x - 1 = 0
x = [ - 5 ± √ (25 + 40 ) ] / 20
x = [ - 5 ± √ (65) ] / 20
x = [ - 5 ± √ 8.06 ] / 20
x = 0.153 , x = - 0.653

x^2+0.5x-0.1=0
=>10x^2 +5x-1=0 multiplying by 10

=>x= {-5 +/- sqrt[25 - 4.10.(-1)]/2.10
= { -5 +/- sqrt[25 + 40]}/20
={-5 +/- sqrt(65)}/20
= {-5 +/- sqrt(65)}/20
= -1/4 +/- 1/20. sqrt(65)<==ANSWER

Easiest thing to do first is get rid of the decimals so it is easier to work with.

x^2 + 0.5x - 0.1 = 0
Multiply the entire thing by 10 you get:

10x^2 + 5x - 1 = 0
Since this cannot be factored, we use the quadratic forumla:
0 = (-b +/- √b^2 - 4ac)/ 2a

Using the forumla you get:
0 = (-5 +/- √(25 - (4*10 * (-1))) / 2(10)
0 = -5 +/- √65 / 20

Divide top and bottom by 5 (common factor) you get:
-1/4 +/- √65

Thus your answer is
-1/4 + √65 and -1/4 - √65