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4t=t2
We move all terms to the left:
4t-(t2)=0
We add all the numbers together, and all the variables
-1t^2+4t=0
a = -1; b = 4; c = 0;
Δ = b2-4ac
Δ = 42-4·(-1)·0
Δ = 16
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{16}=4$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-4}{2*-1}=\frac{-8}{-2} =+4 $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+4}{2*-1}=\frac{0}{-2} =0 $
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