k2+10k=27

Solution for k2+10k=27 equation:

k2+10k=27

We move all terms to the left:

k2+10k-(27)=0

We add all the numbers together, and all the variables

k^2+10k-27=0

a = 1; b = 10; c = -27;
Δ = b2-4ac
Δ = 102-4·1·(-27)
Δ = 208
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{208}=\sqrt{16*13}=\sqrt{16}*\sqrt{13}=4\sqrt{13}$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-4\sqrt{13}}{2*1}=\frac{-10-4\sqrt{13}}{2}
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+4\sqrt{13}}{2*1}=\frac{-10+4\sqrt{13}}{2}
$