6r^2-13-10r=0

Solution for 6r^2-13-10r=0 equation:

6r^2-13-10r=0

a = 6; b = -10; c = -13;
Δ = b2-4ac
Δ = -102-4·6·(-13)
Δ = 412
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:

$\sqrt{\Delta}=\sqrt{412}=\sqrt{4*103}=\sqrt{4}*\sqrt{103}=2\sqrt{103}$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-2\sqrt{103}}{2*6}=\frac{10-2\sqrt{103}}{12}
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+2\sqrt{103}}{2*6}=\frac{10+2\sqrt{103}}{12}
$