Question

How do you solve `3x^2+14x+15=0`?

Answer 1

`x = -5/3` or `x = -3`

Explanation:

`=> 3x^2 + 14x + 15 = 0`

It’s in the form of `ax^2 + bx + c = 0`

where,

• `a = 3`
• `b = 14`
• `c = 15`

Use formula for quadratic equation to find `x`

`x = (-b +- sqrt(b^2 - 4ac))/(2a)`

`x = (-14 +- sqrt(14^2 - (4 × 3 × 15)))/(2 × 3)`

`x = (-14 +- sqrt(196 - 180))/(6)`

`x = (-14 +- sqrt(16))/6`

`x = (-14 +-4)/6`

`x = (-14 + 4)/6 color(white)(....) "or" color(white)(....) x = (-14 - 4)/6`

`x = (-10)/6 color(white)(..........) "or" color(white)(....) x = (-18)/6`

`x = -5/3 color(white)(..........) "or" color(white)(....)x = -3`

Answer 2

I would use the quadratic formula.

Explanation:

The quadratic formula is applicable on equations of the form:

`ax^2+bx+c=0`

And looks like this:

`x= (-b+- sqrt(b^2-4ac))/(2a)`

In our case:

`a=3`
`b=14`
`c=15`

`x= (-14+- sqrt((14)^2-4(3)(15)))/((2)(3))`

`x= (-14+- sqrt(196-180))/(6)`

`x= (-14+- sqrt(16))/(6)`

`x= (-14+- 4)/(6)`

So:

`x_1=(-18/6)=-3`

`x_2=(-10/6)=-5/3`