2r^2+8r+3=-3

Solution for 2r^2+8r+3=-3 equation:

2r^2+8r+3=-3

We move all terms to the left:

2r^2+8r+3-(-3)=0

We add all the numbers together, and all the variables

2r^2+8r+6=0

a = 2; b = 8; c = +6;
Δ = b2-4ac
Δ = 82-4·2·6
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-4}{2*2}=\frac{-12}{4}
=-3
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+4}{2*2}=\frac{-4}{4}
=-1
$