3r(2r-5)=5

Solution for 3r(2r-5)=5 equation:

3r(2r-5)=5

We move all terms to the left:

3r(2r-5)-(5)=0

We multiply parentheses

6r^2-15r-5=0

a = 6; b = -15; c = -5;
Δ = b2-4ac
Δ = -152-4·6·(-5)
Δ = 345
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{-(-15)-\sqrt{345}}{2*6}=\frac{15-\sqrt{345}}{12}
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-15)+\sqrt{345}}{2*6}=\frac{15+\sqrt{345}}{12}
$