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

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

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

We move all terms to the left:

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

We multiply parentheses

-2r^2+8r-10r-(5(2r-6)-8)=0


We calculate terms in parentheses: -(5(2r-6)-8), so:

5(2r-6)-8

We multiply parentheses

10r-30-8

We add all the numbers together, and all the variables

10r-38

Back to the equation:

-(10r-38)


We add all the numbers together, and all the variables

-2r^2-2r-(10r-38)=0

We get rid of parentheses

-2r^2-2r-10r+38=0

We add all the numbers together, and all the variables

-2r^2-12r+38=0

a = -2; b = -12; c = +38;
Δ = b2-4ac
Δ = -122-4·(-2)·38
Δ = 448
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:

$\sqrt{\Delta}=\sqrt{448}=\sqrt{64*7}=\sqrt{64}*\sqrt{7}=8\sqrt{7}$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-8\sqrt{7}}{2*-2}=\frac{12-8\sqrt{7}}{-4}
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+8\sqrt{7}}{2*-2}=\frac{12+8\sqrt{7}}{-4}
$