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w2+29w=0
We add all the numbers together, and all the variables
w^2+29w=0
a = 1; b = 29; c = 0;
Δ = b2-4ac
Δ = 292-4·1·0
Δ = 841
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{841}=29$$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(29)-29}{2*1}=\frac{-58}{2} =-29 $$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(29)+29}{2*1}=\frac{0}{2} =0 $
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