**Question 1: **Solve the given quadratic formula, 2x

^{2}+4x+1=0.

** Solution: **

The given quadratic equation is,

2x^{2}+4x+1=0

We have to find out the discriminant by using the given equation

$\Delta$ =$B^{2}$-4AC

$\Delta$ =$4^{2}$-4×2×1=16-8=8

The roots are given by,

x=$\frac{-B\pm \sqrt{B^{2}-4AC}}{2A}$

x=$\frac{-4\pm \sqrt{8}}{2×2}$

$x_{1}$=$\frac{-4+\sqrt{8}}{4}$ and $x_{2}$=$\frac{-4-\sqrt{8}}{4}$

$x_{1}$=-1+$\frac{1}{\sqrt{2}}$ and $x_{2}$=-1-$\frac{1}{\sqrt{2}}$

$x_{1}$=-1+0.7071=-0.2929 and $x_{2}$=-1-0.7071=-1.7071

**Question 2: **Find the discriminant and roots of the given quadratic equation x

^{2}+6x+3=0?

** Solution: **

The given quadratic equation is,

x^{2}+6x+3=0

We have to find out the discriminant by using the given equation

$\Delta$ =$B^{2}$-4AC

$\Delta$ =$6^{2}$-4×1×3=36-12=24

The roots are given by,

x=$\frac{-B\pm \sqrt{B^{2}-4AC}}{2A}$

x=$\frac{-6\pm \sqrt{24}}{2×1}$

$x_{1}$=-3+$\sqrt{6}$ and $x_{2}$=-3-$\sqrt{6}$

$x_{1}$=-3+2.4494=-0.5505 and $x_{2}$=-3-2.4494=-5.4494