Quadratic? Easy one!

$3x^2+2\sqrt{3}x+2=0$

Comparing the above equation with standard form $ax^2+bx+c=0$ , we will use Vieta's formula for finding the sum.

$\Rightarrow \text{sum} = \dfrac{-b}{a} =\dfrac{-2\sqrt{3}}{3} \approx -1.15$

@Nihar Mahajan provided a simpler, method here another one a bit longer one.

We know that the solution of a quadratic equation is given by $\dfrac{-b\pm\sqrt{b^{2}-4ac}}{2a}$

Here $a=3$ , $b=2\sqrt{3}$ , $c=2$ , substituting these values, we get two solutions for $x$ .

$x=\dfrac{-3+\sqrt{2\sqrt{3}^{2}-4×3×2}}{6}\\ Or, x=\dfrac{-3-\sqrt{2\sqrt{3}^{2}-4×3×2}}{6}$

Therefore, sum of both the values is,

$\dfrac{-2\sqrt{3}+\sqrt{-12}-2\sqrt{3}-\sqrt{-12}}{6}\\ =\dfrac{-4\sqrt{3}}{6}\\=\dfrac{-2\sqrt{3}}{3} \approx \boxed{-1.15}$

A quadratic equation is in the form ax^2 + bx + c ,where the sum of roots is given by

Sum of roots of a quadratic equation= $\frac { -b }{ a }$

$Here, -b=$ $-2\sqrt { 3 }$

$And, a=$ $3$

solving it, we get $\frac { -2\sqrt { 3 } } { 3 }$

which is approxiamately $\boxed { -1.15 }$

a = 3

b = 2sqrt(3)

c =2

Sum of roots = -b/a

Therefore sum of roots = -[2sqrt(3)]/3 which is approximately equal to -1.15