Question

Solve by using the quadratic formula?

`3x^2+4x+10=0`

Answer 1

See a solution process below:

Explanation:

The quadratic formula states:

For `color(red)(a)x^2 + color(blue)(b)x + color(green)(c) = 0`, the values of `x` which are the solutions to the equation are given by:

`x = (-color(blue)(b) +- sqrt(color(blue)(b)^2 - (4color(red)(a)color(green)(c))))/(2 * color(red)(a))`

Substituting:

`color(red)(3)` for `color(red)(a)`

`color(blue)(4)` for `color(blue)(b)`

`color(green)(10)` for `color(green)(c)` gives:

`x = (-color(blue)(4) +- sqrt(color(blue)(4)^2 - (4 * color(red)(3) * color(green)(10))))/(2 * color(red)(3))`

`x = (-color(blue)(4) +- sqrt(16 - 120))/6`

`x = (-color(blue)(4) +- sqrt(-104))/6`

`x = (-color(blue)(4) +- sqrt(4 xx -26))/6`

`x = (-color(blue)(4) +- sqrt(4)sqrt(-26))/6`

`x = (-color(blue)(4) +- 2sqrt(-26))/6`

Answer 2

No real solution.

Explanation:

The quadratic formular is `x= (-b+- sqrt(b^2-4ac))/(2a)` for the equation `color(red)(a)x^2+color(blue)(b)x+color(orange)(c)=0`

Therefore, in your case (`color(red)(3)x^2+color(blue)(4)x+color(orange)(10)=0`)
`a=color(red)(3)`
`b=color(blue)(4)`
`c=color(orange)(10)`

Using the formular, we get:
`x= (-color(blue)(4)+- sqrt(color(blue)(4)^2-4*color(red)(3)*color(orange)(10)))/(2*color(red)(3))`
`x= (-4+- sqrt(16-120))/(6)`
`x=-2/3+-sqrt(color(green)(-104))/6`

Since the radicand (`color(green)(-104)`) is negative, this equation has no real solutions for `x`.