r2+16r=5

Solution for r2+16r=5 equation:

r2+16r=5

We move all terms to the left:

r2+16r-(5)=0

We add all the numbers together, and all the variables

r^2+16r-5=0

a = 1; b = 16; c = -5;
Δ = b2-4ac
Δ = 162-4·1·(-5)
Δ = 276
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:

$\sqrt{\Delta}=\sqrt{276}=\sqrt{4*69}=\sqrt{4}*\sqrt{69}=2\sqrt{69}$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-2\sqrt{69}}{2*1}=\frac{-16-2\sqrt{69}}{2}
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+2\sqrt{69}}{2*1}=\frac{-16+2\sqrt{69}}{2}
$