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4t^2+19t+12=0
a = 4; b = 19; c = +12;
Δ = b2-4ac
Δ = 192-4·4·12
Δ = 169
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{169}=13$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(19)-13}{2*4}=\frac{-32}{8} =-4 $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(19)+13}{2*4}=\frac{-6}{8} =-3/4 $
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