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22x^2+66x=0
a = 22; b = 66; c = 0;
Δ = b2-4ac
Δ = 662-4·22·0
Δ = 4356
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{4356}=66$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(66)-66}{2*22}=\frac{-132}{44} =-3 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(66)+66}{2*22}=\frac{0}{44} =0 $
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