-2r(r+5)=-29

Solution for -2r(r+5)=-29 equation:

-2r(r+5)=-29

We move all terms to the left:

-2r(r+5)-(-29)=0

We add all the numbers together, and all the variables

-2r(r+5)+29=0

We multiply parentheses

-2r^2-10r+29=0

a = -2; b = -10; c = +29;
Δ = b2-4ac
Δ = -102-4·(-2)·29
Δ = 332
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:

$\sqrt{\Delta}=\sqrt{332}=\sqrt{4*83}=\sqrt{4}*\sqrt{83}=2\sqrt{83}$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-2\sqrt{83}}{2*-2}=\frac{10-2\sqrt{83}}{-4}
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+2\sqrt{83}}{2*-2}=\frac{10+2\sqrt{83}}{-4}
$