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3t^2-5t^2+2t=0
We add all the numbers together, and all the variables
-2t^2+2t=0
a = -2; b = 2; c = 0;
Δ = b2-4ac
Δ = 22-4·(-2)·0
Δ = 4
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}$$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{4}=2$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2}{2*-2}=\frac{-4}{-4} =1 $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2}{2*-2}=\frac{0}{-4} =0 $
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