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4t-1=t^2
We move all terms to the left:
4t-1-(t^2)=0
determiningTheFunctionDomain -t^2+4t-1=0
We add all the numbers together, and all the variables
-1t^2+4t-1=0
a = -1; b = 4; c = -1;
Δ = b2-4ac
Δ = 42-4·(-1)·(-1)
Δ = 12
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}$$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{12}=\sqrt{4*3}=\sqrt{4}*\sqrt{3}=2\sqrt{3}$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-2\sqrt{3}}{2*-1}=\frac{-4-2\sqrt{3}}{-2} $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+2\sqrt{3}}{2*-1}=\frac{-4+2\sqrt{3}}{-2} $
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