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