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4t^2=4
We move all terms to the left:
4t^2-(4)=0
a = 4; b = 0; c = -4;
Δ = b2-4ac
Δ = 02-4·4·(-4)
Δ = 64
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{64}=8$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-8}{2*4}=\frac{-8}{8} =-1 $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+8}{2*4}=\frac{8}{8} =1 $
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