k2+17k+24=0

Solution for k2+17k+24=0 equation:

k2+17k+24=0

We add all the numbers together, and all the variables

k^2+17k+24=0

a = 1; b = 17; c = +24;
Δ = b2-4ac
Δ = 172-4·1·24
Δ = 193
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(17)-\sqrt{193}}{2*1}=\frac{-17-\sqrt{193}}{2}
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(17)+\sqrt{193}}{2*1}=\frac{-17+\sqrt{193}}{2}
$