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t2+5t=36
We move all terms to the left:
t2+5t-(36)=0
We add all the numbers together, and all the variables
t^2+5t-36=0
a = 1; b = 5; c = -36;
Δ = b2-4ac
Δ = 52-4·1·(-36)
Δ = 169
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{169}=13$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(5)-13}{2*1}=\frac{-18}{2} =-9 $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+13}{2*1}=\frac{8}{2} =4 $
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