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t2=5t
We move all terms to the left:
t2-(5t)=0
We add all the numbers together, and all the variables
t^2-5t=0
a = 1; b = -5; c = 0;
Δ = b2-4ac
Δ = -52-4·1·0
Δ = 25
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{25}=5$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-5)-5}{2*1}=\frac{0}{2} =0 $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+5}{2*1}=\frac{10}{2} =5 $
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