r2+12r+-41=0

Solution for r2+12r+-41=0 equation:

r2+12r+-41=0

We add all the numbers together, and all the variables

r^2+12r=0

a = 1; b = 12; c = 0;
Δ = b2-4ac
Δ = 122-4·1·0
Δ = 144
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{144}=12$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-12}{2*1}=\frac{-24}{2}
=-12
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+12}{2*1}=\frac{0}{2}
=0
$