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t=-16t^2=64t
We move all terms to the left:
t-(-16t^2)=0
We get rid of parentheses
16t^2+t=0
a = 16; b = 1; c = 0;
Δ = b2-4ac
Δ = 12-4·16·0
Δ = 1
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{1}=1$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-1}{2*16}=\frac{-2}{32} =-1/16 $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+1}{2*16}=\frac{0}{32} =0 $
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