r^2+3r+6=4

Solution for r^2+3r+6=4 equation:

r^2+3r+6=4

We move all terms to the left:

r^2+3r+6-(4)=0

We add all the numbers together, and all the variables

r^2+3r+2=0

a = 1; b = 3; c = +2;
Δ = b2-4ac
Δ = 32-4·1·2
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

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