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3t^2+16t+20=0
a = 3; b = 16; c = +20;
Δ = b2-4ac
Δ = 162-4·3·20
Δ = 16
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{16}=4$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-4}{2*3}=\frac{-20}{6} =-3+1/3 $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+4}{2*3}=\frac{-12}{6} =-2 $
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