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k^2-12k=-32
We move all terms to the left:
k^2-12k-(-32)=0
We add all the numbers together, and all the variables
k^2-12k+32=0
a = 1; b = -12; c = +32;
Δ = b2-4ac
Δ = -122-4·1·32
Δ = 16
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{16}=4$$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-4}{2*1}=\frac{8}{2} =4 $$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+4}{2*1}=\frac{16}{2} =8 $
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