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16t^2+36t=0
a = 16; b = 36; c = 0;
Δ = b2-4ac
Δ = 362-4·16·0
Δ = 1296
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{1296}=36$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(36)-36}{2*16}=\frac{-72}{32} =-2+1/4 $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(36)+36}{2*16}=\frac{0}{32} =0 $
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