-4(r2r+2)-5=-13-8r

Solution for -4(r2r+2)-5=-13-8r equation:

-4(r2r+2)-5=-13-8r

We move all terms to the left:

-4(r2r+2)-5-(-13-8r)=0

We add all the numbers together, and all the variables

-4(r2r+2)-(-8r-13)-5=0

We multiply parentheses

-4r^2-(-8r-13)-8-5=0

We get rid of parentheses

-4r^2+8r+13-8-5=0

We add all the numbers together, and all the variables

-4r^2+8r=0

a = -4; b = 8; c = 0;
Δ = b2-4ac
Δ = 82-4·(-4)·0
Δ = 64
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{64}=8$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-8}{2*-4}=\frac{-16}{-8}
=+2
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+8}{2*-4}=\frac{0}{-8}
=0
$