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4t^2-6t-18=0
a = 4; b = -6; c = -18;
Δ = b2-4ac
Δ = -62-4·4·(-18)
Δ = 324
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{324}=18$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-6)-18}{2*4}=\frac{-12}{8} =-1+1/2 $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-6)+18}{2*4}=\frac{24}{8} =3 $
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