r^2+14r+29=5

Solution for r^2+14r+29=5 equation:

r^2+14r+29=5

We move all terms to the left:

r^2+14r+29-(5)=0

We add all the numbers together, and all the variables

r^2+14r+24=0

a = 1; b = 14; c = +24;
Δ = b2-4ac
Δ = 142-4·1·24
Δ = 100
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{100}=10$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(14)-10}{2*1}=\frac{-24}{2}
=-12
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(14)+10}{2*1}=\frac{-4}{2}
=-2
$