-8r(2r-5)=6+r

Solution for -8r(2r-5)=6+r equation:

-8r(2r-5)=6+r

We move all terms to the left:

-8r(2r-5)-(6+r)=0

We add all the numbers together, and all the variables

-8r(2r-5)-(r+6)=0

We multiply parentheses

-16r^2+40r-(r+6)=0

We get rid of parentheses

-16r^2+40r-r-6=0

We add all the numbers together, and all the variables

-16r^2+39r-6=0

a = -16; b = 39; c = -6;
Δ = b2-4ac
Δ = 392-4·(-16)·(-6)
Δ = 1137
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{-(39)-\sqrt{1137}}{2*-16}=\frac{-39-\sqrt{1137}}{-32}
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(39)+\sqrt{1137}}{2*-16}=\frac{-39+\sqrt{1137}}{-32}
$