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9-4t^2=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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