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d^2+46d=0
a = 1; b = 46; c = 0;
Δ = b2-4ac
Δ = 462-4·1·0
Δ = 2116
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{2116}=46$$d_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(46)-46}{2*1}=\frac{-92}{2} =-46 $$d_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(46)+46}{2*1}=\frac{0}{2} =0 $
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