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36t^2-25=0
a = 36; b = 0; c = -25;
Δ = b2-4ac
Δ = 02-4·36·(-25)
Δ = 3600
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{3600}=60$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-60}{2*36}=\frac{-60}{72} =-5/6 $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+60}{2*36}=\frac{60}{72} =5/6 $
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