# How do you find the zeros of  y = 2x^2 + 4x -1  using the quadratic formula?

Dec 13, 2017

$- 1 \pm \frac{\sqrt{6}}{2}$

#### Explanation:

This quadratic function is written in "standard form", or $a {x}^{2} + b x + c$.
The quadratic formula uses the numbers a, b, and c like this:
$\frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$
comparing the original function to the general, standard form function above, we see that a=2, b=4, &c=-1.

Plug these in the quadratic formula:
$\frac{- 4 \pm \sqrt{{4}^{2} - 4 \left(2\right) \left(- 1\right)}}{2 \cdot 2} = \frac{- 4 \pm \sqrt{16 - \left(- 8\right)}}{4}$
$\frac{- 4 \pm \sqrt{24}}{4}$
To simplify that $\sqrt{24}$, pull out the largest perfect square that goes in it. Since 4 divides evenly into 24, and neither 9 nor 16 do, break it down in terms of 4: $\sqrt{24} = \sqrt{4 \cdot 6}$
You can take out $\sqrt{4} = 2$:

$\sqrt{4 \cdot 6} = 2 \sqrt{6}$
From here, divide through by 4 to get $- 1 \pm \frac{\sqrt{6}}{2}$

...and remember, quadratic functions always have two solutions!