k2+13k=-40

Solution for k2+13k=-40 equation:

k2+13k=-40

We move all terms to the left:

k2+13k-(-40)=0

We add all the numbers together, and all the variables

k^2+13k+40=0

a = 1; b = 13; c = +40;
Δ = b2-4ac
Δ = 132-4·1·40
Δ = 9
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{9}=3$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(13)-3}{2*1}=\frac{-16}{2}
=-8
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(13)+3}{2*1}=\frac{-10}{2}
=-5
$