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4t^2+35t-9=0
a = 4; b = 35; c = -9;
Δ = b2-4ac
Δ = 352-4·4·(-9)
Δ = 1369
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{1369}=37$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(35)-37}{2*4}=\frac{-72}{8} =-9 $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(35)+37}{2*4}=\frac{2}{8} =1/4 $
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