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16t^2-64t+48=0
a = 16; b = -64; c = +48;
Δ = b2-4ac
Δ = -642-4·16·48
Δ = 1024
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{1024}=32$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-64)-32}{2*16}=\frac{32}{32} =1 $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-64)+32}{2*16}=\frac{96}{32} =3 $
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