How do you solve 2x²+5x+2=0 in the quadratic formula ?

Question Number 1 :

For this equation 2*x^2 + 5*x + 2 = 0 , answer the following questions :

A. Find the roots using Quadratic Formula !

Answer Number 1 :

The equation 2*x^2 + 5*x + 2 = 0 is already in a*x^2+b*x+c=0 form.

So we can imply that the value of a = 2, b = 5, c = 2.

1A. Find the roots using Quadratic Formula !

Use abc formula and you get either

x1 = (-b+sqrt(b^2-4*a*c))/(2*a) or x2 = (-b-sqrt(b^2-4*a*c))/(2*a)

Since a = 2, b = 5 and c = 2,

then the value a,b and c in the abc formula, can be subtituted.

So x1 = (-(5) + sqrt( (5)^2 - 4 * (2)*(2)))/(2*2) and x2 = (-(5) - sqrt( (5)^2 - 4 * (2)*(2)))/(2*2)

Which is the same as x1 = ( -5 + sqrt( 25-16))/(4) and x2 = ( -5 - sqrt( 25-16))/(4)

Which is the same as x1 = ( -5 + sqrt( 9))/(4) and x2 = ( -5 - sqrt( 9))/(4)

We got x1 = ( -5 + 3 )/(4) and x2 = ( -5 - 3 )/(4)

So we got the answers as x1 = -0.5 and x2 = -2

A is the number before the x^2

B is the number before the x and

C is the number all by its lonesome

So for your equation:

A=2

B=5

C=2

Now that you know that you can simply plug these numbers into the equation and get the answer. Remember there is a + or - before the square root, so you will likely get two answers

-5+√(5*5)-4*(2)*(2) / 2*2 = -1/2

-5-√(5*5)-4*(2)*(2) / 2*2 = -2

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2x²+5x+2=0

a is you x²

b is your x

and c is your one without any x's

a=+2

b=+5

c=+2

so ...

-5+/-√25-(4*2*2) all divided by 4

-5+√25-16= -5+√9 divided by 4

which is

-5+3=-2

-2/4= -0.5

and then

-5-√9= -5-3= -8

-8/4= -2

....................................................................................................................

by the way the quadratic equation is -b(+/-)√b²-4ac all divided by 2a

I really dont know, but remember Please Xuse my dear aunt sally.