3r2+3=199-4

Solution for 3r2+3=199-4 equation:

3r^2+3=199-4

We move all terms to the left:

3r^2+3-(199-4)=0

We add all the numbers together, and all the variables

3r^2+3-195=0

We add all the numbers together, and all the variables

3r^2-192=0

a = 3; b = 0; c = -192;
Δ = b2-4ac
Δ = 02-4·3·(-192)
Δ = 2304
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{2304}=48$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-48}{2*3}=\frac{-48}{6}
=-8
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+48}{2*3}=\frac{48}{6}
=8
$