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36=t2
We move all terms to the left:
36-(t2)=0
We add all the numbers together, and all the variables
-1t^2+36=0
a = -1; b = 0; c = +36;
Δ = b2-4ac
Δ = 02-4·(-1)·36
Δ = 144
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{144}=12$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-12}{2*-1}=\frac{-12}{-2} =+6 $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+12}{2*-1}=\frac{12}{-2} =-6 $
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