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2t^2-25t+12=0
a = 2; b = -25; c = +12;
Δ = b2-4ac
Δ = -252-4·2·12
Δ = 529
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{529}=23$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-25)-23}{2*2}=\frac{2}{4} =1/2 $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-25)+23}{2*2}=\frac{48}{4} =12 $
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