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5t^2+2.5t-4=0
a = 5; b = 2.5; c = -4;
Δ = b2-4ac
Δ = 2.52-4·5·(-4)
Δ = 86.25
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}$$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2.5)-\sqrt{86.25}}{2*5}=\frac{-2.5-\sqrt{86.25}}{10} $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2.5)+\sqrt{86.25}}{2*5}=\frac{-2.5+\sqrt{86.25}}{10} $
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