20(r+5)(-40=10)2r+6

Solution for 20(r+5)(-40=10)2r+6 equation:

20(r+5)(-40=10)2r+6

We move all terms to the left:

20(r+5)(-40-(10)2r+6)=0

We add all the numbers together, and all the variables

20(r+5)(-102r-34)=0

We multiply parentheses ..

20(-102r^2-34r-510r-170)=0

We multiply parentheses

-2040r^2-680r-10200r-3400=0

We add all the numbers together, and all the variables

-2040r^2-10880r-3400=0

a = -2040; b = -10880; c = -3400;
Δ = b2-4ac
Δ = -108802-4·(-2040)·(-3400)
Δ = 90630400
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{90630400}=9520$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10880)-9520}{2*-2040}=\frac{1360}{-4080}
=-1/3
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10880)+9520}{2*-2040}=\frac{20400}{-4080}
=-5
$