Question

Use the quadratic formula to find both solutions to the quadratic equation given below. 2x^2+3x-5=0

Answer 1

B and C

Explanation:

Answer 2

`x =(-b \pm √(b^2 -4ac))/(2a)`

Where a = 2 , b= 3, c= -5, replacing we have this:

`x =(-3 \pm √((-3)^2 -4(2)(-5)))/(2*2)`

And simplifying we got:

`x = (-3 \pm √(49))/(4)`

And the two solutions are:

`x_1 = (-3+7)/(4)= 1`

`x_2 = (-3-7)/(4)= -(5)/(2)`

And the correct options are:

B and C

Explanation:

We have the following equation given:

`2x^2 +3x -5=0`

And if we use the quadratic formula given by:

`x =(-b \pm √(b^2 -4ac))/(2a)`

Where a = 2 , b= 3, c= -5, replacing we have this:

`x =(-3 \pm √((-3)^2 -4(2)(-5)))/(2*2)`

And simplifying we got:

`x = (-3 \pm √(49))/(4)`

And the two solutions are:

`x_1 = (-3+7)/(4)= 1`

`x_2 = (-3-7)/(4)= -(5)/(2)`

And the correct options are:

B and C