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4t^2-9=0
a = 4; b = 0; c = -9;
Δ = b2-4ac
Δ = 02-4·4·(-9)
Δ = 144
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{144}=12$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-12}{2*4}=\frac{-12}{8} =-1+1/2 $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+12}{2*4}=\frac{12}{8} =1+1/2 $
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