r2+r/3=410

Solution for r2+r/3=410 equation:

r2+r/3=410

We move all terms to the left:

r2+r/3-(410)=0

We add all the numbers together, and all the variables

r^2+r/3-410=0

We multiply all the terms by the denominator


r^2*3+r-410*3=0

We add all the numbers together, and all the variables

r^2*3+r-1230=0

Wy multiply elements

3r^2+r-1230=0

a = 3; b = 1; c = -1230;
Δ = b2-4ac
Δ = 12-4·3·(-1230)
Δ = 14761
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-\sqrt{14761}}{2*3}=\frac{-1-\sqrt{14761}}{6}
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{14761}}{2*3}=\frac{-1+\sqrt{14761}}{6}
$