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0=-16t^2+11t+3
We move all terms to the left:
0-(-16t^2+11t+3)=0
We add all the numbers together, and all the variables
-(-16t^2+11t+3)=0
We get rid of parentheses
16t^2-11t-3=0
a = 16; b = -11; c = -3;
Δ = b2-4ac
Δ = -112-4·16·(-3)
Δ = 313
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}$$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-11)-\sqrt{313}}{2*16}=\frac{11-\sqrt{313}}{32} $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-11)+\sqrt{313}}{2*16}=\frac{11+\sqrt{313}}{32} $
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