5(3r-10)r=4

Solution for 5(3r-10)r=4 equation:

5(3r-10)r=4

We move all terms to the left:

5(3r-10)r-(4)=0

We multiply parentheses

15r^2-50r-4=0

a = 15; b = -50; c = -4;
Δ = b2-4ac
Δ = -502-4·15·(-4)
Δ = 2740
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:

$\sqrt{\Delta}=\sqrt{2740}=\sqrt{4*685}=\sqrt{4}*\sqrt{685}=2\sqrt{685}$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-50)-2\sqrt{685}}{2*15}=\frac{50-2\sqrt{685}}{30}
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-50)+2\sqrt{685}}{2*15}=\frac{50+2\sqrt{685}}{30}
$