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0=240-16t^2
We move all terms to the left:
0-(240-16t^2)=0
We add all the numbers together, and all the variables
-(240-16t^2)=0
We get rid of parentheses
16t^2-240=0
a = 16; b = 0; c = -240;
Δ = b2-4ac
Δ = 02-4·16·(-240)
Δ = 15360
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}$$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{15360}=\sqrt{1024*15}=\sqrt{1024}*\sqrt{15}=32\sqrt{15}$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-32\sqrt{15}}{2*16}=\frac{0-32\sqrt{15}}{32} =-\frac{32\sqrt{15}}{32} =-\sqrt{15} $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+32\sqrt{15}}{2*16}=\frac{0+32\sqrt{15}}{32} =\frac{32\sqrt{15}}{32} =\sqrt{15} $
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