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h^2+21h=0
a = 1; b = 21; c = 0;
Δ = b2-4ac
Δ = 212-4·1·0
Δ = 441
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$h_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$h_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{441}=21$$h_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(21)-21}{2*1}=\frac{-42}{2} =-21 $$h_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(21)+21}{2*1}=\frac{0}{2} =0 $
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