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12t^2+25t+12=0
a = 12; b = 25; c = +12;
Δ = b2-4ac
Δ = 252-4·12·12
Δ = 49
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{49}=7$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(25)-7}{2*12}=\frac{-32}{24} =-1+1/3 $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(25)+7}{2*12}=\frac{-18}{24} =-3/4 $
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