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(^2+4+5)D=0
We multiply parentheses
D^2+4D+5D=0
We add all the numbers together, and all the variables
D^2+9D=0
a = 1; b = 9; c = 0;
Δ = b2-4ac
Δ = 92-4·1·0
Δ = 81
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$D_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$D_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{81}=9$$D_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(9)-9}{2*1}=\frac{-18}{2} =-9 $$D_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(9)+9}{2*1}=\frac{0}{2} =0 $
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