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0=16t^2+160t+6
We move all terms to the left:
0-(16t^2+160t+6)=0
We add all the numbers together, and all the variables
-(16t^2+160t+6)=0
We get rid of parentheses
-16t^2-160t-6=0
a = -16; b = -160; c = -6;
Δ = b2-4ac
Δ = -1602-4·(-16)·(-6)
Δ = 25216
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}$$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{25216}=\sqrt{64*394}=\sqrt{64}*\sqrt{394}=8\sqrt{394}$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-160)-8\sqrt{394}}{2*-16}=\frac{160-8\sqrt{394}}{-32} $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-160)+8\sqrt{394}}{2*-16}=\frac{160+8\sqrt{394}}{-32} $
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