# How do you find the zeros for this equation? y=x (to the power of 2) +3x-5

### 1 Answer | Add Yours

To solve for the zeros, set y = 0 and you'll have:

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

Then use quadratic formula: `x = (-b+-sqrt(b^2-4ac))/(2a)`

The quadratic equation above is in the form `ax^2 + bx + c = 0.`

You can now apply the formula.

a = 1

b = 3

c = -5

So, `x = (-3+-sqrt(3^2-4(1)(-5)))/((2)(1))`

`x = (-3+-sqrt(9+20))/2`

`x = (-3 + sqrt(29))/2`

`x = 1.19 (rounded to two decimal position)`

`x=(-3-sqrt(29))/2`

`x = -4.19(rounded to two decimal position)`

Since it is a quadratic equation (with 2 as the highest exponent), you should have 2 values for x.

The roots are 1.19 and - 4.19.