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0=-120t^2+-240t+12
We move all terms to the left:
0-(-120t^2+-240t+12)=0
We add all the numbers together, and all the variables
-(-120t^2+-240t+12)=0
We use the square of the difference formula
-(-120t^2-240t+12)=0
We get rid of parentheses
120t^2+240t-12=0
a = 120; b = 240; c = -12;
Δ = b2-4ac
Δ = 2402-4·120·(-12)
Δ = 63360
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}$$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{63360}=\sqrt{576*110}=\sqrt{576}*\sqrt{110}=24\sqrt{110}$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(240)-24\sqrt{110}}{2*120}=\frac{-240-24\sqrt{110}}{240} $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(240)+24\sqrt{110}}{2*120}=\frac{-240+24\sqrt{110}}{240} $
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