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h2+12h+27=0
We add all the numbers together, and all the variables
h^2+12h+27=0
a = 1; b = 12; c = +27;
Δ = b2-4ac
Δ = 122-4·1·27
Δ = 36
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$h_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$h_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{36}=6$$h_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-6}{2*1}=\frac{-18}{2} =-9 $$h_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+6}{2*1}=\frac{-6}{2} =-3 $
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