Question

How do I solve


`2x^(2)+5x+2=0`

Answer 1

There are two solutions.

`x_1 = -0.5\\x_2 = -2`

Explanation:

Step 1

Use the formula of ax^2 + bx + c = 0 to find the coefficients.

`a = 2\\b = 5\\c = 2`

Step 2

`x=( (-b+/-sqrt(b^2-4ac)))/((2a))\\x=( (-5 +/- sqrt(5^2-4*2*2)))/((2*2))\\x=((-5 +/- sqrt(25-4*2*2)))/((2*2))\\x=((-5 +/- sqrt(25-8*2)))/((2*2))\\x=((-5 +/- sqrt(9)))/((2*2))`

`x = (-5 +/- √(9) )/(4)`

Step 3

`9 = 3^2\\√(9) = √(3*3)\\√(3*3) = √(3^2)\\√(3^2) = 3`

Step 4

`x_1=((-5+3))/(4) -- > x_2=((-5-3))/(4)\\x_1=((-5+3))/(4)\\x_1=(-2)/(4)\\x_1=-0,5`

`x_2=((-5-3))/(4)\\x_2=((-8))/(4)`

`x_2=-2`

Hope this helps.

Answer 2

see the attachment...

hope it helps...!!!