k2-12k-57=6

Solution for k2-12k-57=6 equation:

k2-12k-57=6

We move all terms to the left:

k2-12k-57-(6)=0

We add all the numbers together, and all the variables

k^2-12k-63=0

a = 1; b = -12; c = -63;
Δ = b2-4ac
Δ = -122-4·1·(-63)
Δ = 396
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:

$\sqrt{\Delta}=\sqrt{396}=\sqrt{36*11}=\sqrt{36}*\sqrt{11}=6\sqrt{11}$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-6\sqrt{11}}{2*1}=\frac{12-6\sqrt{11}}{2}
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+6\sqrt{11}}{2*1}=\frac{12+6\sqrt{11}}{2}
$