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t2+22t=0
We add all the numbers together, and all the variables
t^2+22t=0
a = 1; b = 22; c = 0;
Δ = b2-4ac
Δ = 222-4·1·0
Δ = 484
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{484}=22$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(22)-22}{2*1}=\frac{-44}{2} =-22 $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(22)+22}{2*1}=\frac{0}{2} =0 $
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