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80=3t^2-30t+135
We move all terms to the left:
80-(3t^2-30t+135)=0
We get rid of parentheses
-3t^2+30t-135+80=0
We add all the numbers together, and all the variables
-3t^2+30t-55=0
a = -3; b = 30; c = -55;
Δ = b2-4ac
Δ = 302-4·(-3)·(-55)
Δ = 240
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}$$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{240}=\sqrt{16*15}=\sqrt{16}*\sqrt{15}=4\sqrt{15}$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(30)-4\sqrt{15}}{2*-3}=\frac{-30-4\sqrt{15}}{-6} $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(30)+4\sqrt{15}}{2*-3}=\frac{-30+4\sqrt{15}}{-6} $
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