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b2+18b=0
We add all the numbers together, and all the variables
b^2+18b=0
a = 1; b = 18; c = 0;
Δ = b2-4ac
Δ = 182-4·1·0
Δ = 324
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{324}=18$$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(18)-18}{2*1}=\frac{-36}{2} =-18 $$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(18)+18}{2*1}=\frac{0}{2} =0 $
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