r2+4r+3=0

Solution for r2+4r+3=0 equation:

r2+4r+3=0

We add all the numbers together, and all the variables

r^2+4r+3=0

a = 1; b = 4; c = +3;
Δ = b2-4ac
Δ = 42-4·1·3
Δ = 4
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

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