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24=2+40t-16t^2
We move all terms to the left:
24-(2+40t-16t^2)=0
We get rid of parentheses
16t^2-40t-2+24=0
We add all the numbers together, and all the variables
16t^2-40t+22=0
a = 16; b = -40; c = +22;
Δ = b2-4ac
Δ = -402-4·16·22
Δ = 192
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}$$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{192}=\sqrt{64*3}=\sqrt{64}*\sqrt{3}=8\sqrt{3}$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-40)-8\sqrt{3}}{2*16}=\frac{40-8\sqrt{3}}{32} $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-40)+8\sqrt{3}}{2*16}=\frac{40+8\sqrt{3}}{32} $
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