-5(1-5r)=6(-5+5r)r

Solution for -5(1-5r)=6(-5+5r)r equation:

-5(1-5r)=6(-5+5r)r

We move all terms to the left:

-5(1-5r)-(6(-5+5r)r)=0

We add all the numbers together, and all the variables

-5(-5r+1)-(6(5r-5)r)=0

We multiply parentheses

25r-(6(5r-5)r)-5=0


We calculate terms in parentheses: -(6(5r-5)r), so:

6(5r-5)r

We multiply parentheses

30r^2-30r

Back to the equation:

-(30r^2-30r)


We get rid of parentheses

-30r^2+25r+30r-5=0

We add all the numbers together, and all the variables

-30r^2+55r-5=0

a = -30; b = 55; c = -5;
Δ = b2-4ac
Δ = 552-4·(-30)·(-5)
Δ = 2425
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:

$\sqrt{\Delta}=\sqrt{2425}=\sqrt{25*97}=\sqrt{25}*\sqrt{97}=5\sqrt{97}$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(55)-5\sqrt{97}}{2*-30}=\frac{-55-5\sqrt{97}}{-60}
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(55)+5\sqrt{97}}{2*-30}=\frac{-55+5\sqrt{97}}{-60}
$