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5+8t+3t^2=0
a = 3; b = 8; c = +5;
Δ = b2-4ac
Δ = 82-4·3·5
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$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-2}{2*3}=\frac{-10}{6} =-1+2/3 $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+2}{2*3}=\frac{-6}{6} =-1 $
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