-10(r+12)-4=-10(r+11)r

Solution for -10(r+12)-4=-10(r+11)r equation:

-10(r+12)-4=-10(r+11)r

We move all terms to the left:

-10(r+12)-4-(-10(r+11)r)=0

We multiply parentheses

-10r-(-10(r+11)r)-120-4=0


We calculate terms in parentheses: -(-10(r+11)r), so:

-10(r+11)r

We multiply parentheses

-10r^2-110r

Back to the equation:

-(-10r^2-110r)


We add all the numbers together, and all the variables

-(-10r^2-110r)-10r-124=0

We get rid of parentheses

10r^2+110r-10r-124=0

We add all the numbers together, and all the variables

10r^2+100r-124=0

a = 10; b = 100; c = -124;
Δ = b2-4ac
Δ = 1002-4·10·(-124)
Δ = 14960
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:

$\sqrt{\Delta}=\sqrt{14960}=\sqrt{16*935}=\sqrt{16}*\sqrt{935}=4\sqrt{935}$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(100)-4\sqrt{935}}{2*10}=\frac{-100-4\sqrt{935}}{20}
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(100)+4\sqrt{935}}{2*10}=\frac{-100+4\sqrt{935}}{20}
$