K(x)=2x^2+7x+3

Solution for K(x)=2x^2+7x+3 equation:

(K)=2K^2+7K+3

We move all terms to the left:

(K)-(2K^2+7K+3)=0

We get rid of parentheses

-2K^2+K-7K-3=0

We add all the numbers together, and all the variables

-2K^2-6K-3=0

a = -2; b = -6; c = -3;
Δ = b2-4ac
Δ = -62-4·(-2)·(-3)
Δ = 12
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$K_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$K_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{12}=\sqrt{4*3}=\sqrt{4}*\sqrt{3}=2\sqrt{3}$
$K_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-6)-2\sqrt{3}}{2*-2}=\frac{6-2\sqrt{3}}{-4}
$
$K_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-6)+2\sqrt{3}}{2*-2}=\frac{6+2\sqrt{3}}{-4}
$