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t2-5t=-4
We move all terms to the left:
t2-5t-(-4)=0
We add all the numbers together, and all the variables
t^2-5t+4=0
a = 1; b = -5; c = +4;
Δ = b2-4ac
Δ = -52-4·1·4
Δ = 9
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{9}=3$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-5)-3}{2*1}=\frac{2}{2} =1 $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+3}{2*1}=\frac{8}{2} =4 $
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