r2+6r+3=0

Solution for r2+6r+3=0 equation:

r2+6r+3=0

We add all the numbers together, and all the variables

r^2+6r+3=0

a = 1; b = 6; c = +3;
Δ = b2-4ac
Δ = 62-4·1·3
Δ = 24
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{24}=\sqrt{4*6}=\sqrt{4}*\sqrt{6}=2\sqrt{6}$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-2\sqrt{6}}{2*1}=\frac{-6-2\sqrt{6}}{2}
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+2\sqrt{6}}{2*1}=\frac{-6+2\sqrt{6}}{2}
$