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(^3-^2-11+3)D=0
We multiply parentheses
D^2+D^2-11D+3D=0
We add all the numbers together, and all the variables
2D^2-8D=0
a = 2; b = -8; c = 0;
Δ = b2-4ac
Δ = -82-4·2·0
Δ = 64
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{64}=8$$D_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-8)-8}{2*2}=\frac{0}{4} =0 $$D_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-8)+8}{2*2}=\frac{16}{4} =4 $
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