3r^2=-2r+13

Solution for 3r^2=-2r+13 equation:

3r^2=-2r+13

We move all terms to the left:

3r^2-(-2r+13)=0

We get rid of parentheses

3r^2+2r-13=0

a = 3; b = 2; c = -13;
Δ = b2-4ac
Δ = 22-4·3·(-13)
Δ = 160
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{160}=\sqrt{16*10}=\sqrt{16}*\sqrt{10}=4\sqrt{10}$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-4\sqrt{10}}{2*3}=\frac{-2-4\sqrt{10}}{6}
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+4\sqrt{10}}{2*3}=\frac{-2+4\sqrt{10}}{6}
$