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h^2+11h-88=-88
We move all terms to the left:
h^2+11h-88-(-88)=0
We add all the numbers together, and all the variables
h^2+11h=0
a = 1; b = 11; c = 0;
Δ = b2-4ac
Δ = 112-4·1·0
Δ = 121
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{121}=11$$h_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(11)-11}{2*1}=\frac{-22}{2} =-11 $$h_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(11)+11}{2*1}=\frac{0}{2} =0 $
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