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60=-5t^2+40t+40
We move all terms to the left:
60-(-5t^2+40t+40)=0
We get rid of parentheses
5t^2-40t-40+60=0
We add all the numbers together, and all the variables
5t^2-40t+20=0
a = 5; b = -40; c = +20;
Δ = b2-4ac
Δ = -402-4·5·20
Δ = 1200
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}$$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{1200}=\sqrt{400*3}=\sqrt{400}*\sqrt{3}=20\sqrt{3}$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-40)-20\sqrt{3}}{2*5}=\frac{40-20\sqrt{3}}{10} $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-40)+20\sqrt{3}}{2*5}=\frac{40+20\sqrt{3}}{10} $
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