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4t^2+13t+10=0
a = 4; b = 13; c = +10;
Δ = b2-4ac
Δ = 132-4·4·10
Δ = 9
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{9}=3$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(13)-3}{2*4}=\frac{-16}{8} =-2 $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(13)+3}{2*4}=\frac{-10}{8} =-1+1/4 $
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