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(t)=-16t^2+32t+5
We move all terms to the left:
(t)-(-16t^2+32t+5)=0
We get rid of parentheses
16t^2-32t+t-5=0
We add all the numbers together, and all the variables
16t^2-31t-5=0
a = 16; b = -31; c = -5;
Δ = b2-4ac
Δ = -312-4·16·(-5)
Δ = 1281
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{-(-31)-\sqrt{1281}}{2*16}=\frac{31-\sqrt{1281}}{32} $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-31)+\sqrt{1281}}{2*16}=\frac{31+\sqrt{1281}}{32} $
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