k^2+6k-35=-2

Solution for k^2+6k-35=-2 equation:

k^2+6k-35=-2

We move all terms to the left:

k^2+6k-35-(-2)=0

We add all the numbers together, and all the variables

k^2+6k-33=0

a = 1; b = 6; c = -33;
Δ = b2-4ac
Δ = 62-4·1·(-33)
Δ = 168
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:

$\sqrt{\Delta}=\sqrt{168}=\sqrt{4*42}=\sqrt{4}*\sqrt{42}=2\sqrt{42}$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-2\sqrt{42}}{2*1}=\frac{-6-2\sqrt{42}}{2}
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+2\sqrt{42}}{2*1}=\frac{-6+2\sqrt{42}}{2}
$