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4t^2+12t+5=0
a = 4; b = 12; c = +5;
Δ = b2-4ac
Δ = 122-4·4·5
Δ = 64
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{64}=8$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-8}{2*4}=\frac{-20}{8} =-2+1/2 $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+8}{2*4}=\frac{-4}{8} =-1/2 $
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