Question

How do you use the quadratic formula to solve `3x ^ { 2} + 3x - 5= 0`?

Answer 1

`=> x = { ( -3 + sqrt(69) ) / (6) , ( -3 - sqrt(69) ) / (6) }`

Or approximately

`=> x approx { 0.884 , -1.884 }`

Explanation:

The quadratic is `ax^2 + bx +c = 0`

and the formula is: `x =( -b pm sqrt(b^2 - 4ac ) ) / (2a )`

In this case `a = 3` , `b = 3` and `c = -5`

`=> x = ( -3 pm sqrt( 3^2 - (4*3*(-5) ) ) ) / (2*3)`

`=> x = ( -3 pm sqrt(69) ) / (6)`

`=> x = { ( -3 + sqrt(69) ) / (6) , ( -3 - sqrt(69) ) / (6) }`

Or approximately

`=> x approx { 0.884 , -1.884 }`

Answer 2

`x=(−3+sqrt69)/(6)=0.88`
or
`x=(−3-sqrt69)/(6)=-1.88`

Explanation:

The equation `3x^2=3x-5=0` is written in the form `y=ax^2+bx+c`, so `a=3, b=3, c=-5`

The quadratic formula is `x=(−b±sqrt(b^2−4ac))/(2a)`

Substitute the values of a, b and c into the formula

`x=(−(3)±sqrt(3^2−4(3xx-5)))/(2(3))`
`x=(−3±sqrt(9+60))/(6)`
`x=(−3±13)/(6)`

`x=(−3+sqrt69)/(6)=0.88`
or
`x=(−3-sqrt69)/(6)=-1.88`