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(^2+3+2)D=0
We multiply parentheses
D^2+3D+2D=0
We add all the numbers together, and all the variables
D^2+5D=0
a = 1; b = 5; c = 0;
Δ = b2-4ac
Δ = 52-4·1·0
Δ = 25
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{25}=5$$D_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(5)-5}{2*1}=\frac{-10}{2} =-5 $$D_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+5}{2*1}=\frac{0}{2} =0 $
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