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5t^2+12t+7=0
a = 5; b = 12; c = +7;
Δ = b2-4ac
Δ = 122-4·5·7
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$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-2}{2*5}=\frac{-14}{10} =-1+2/5 $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+2}{2*5}=\frac{-10}{10} =-1 $
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