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6t^2-11t=35
We move all terms to the left:
6t^2-11t-(35)=0
a = 6; b = -11; c = -35;
Δ = b2-4ac
Δ = -112-4·6·(-35)
Δ = 961
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{961}=31$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-11)-31}{2*6}=\frac{-20}{12} =-1+2/3 $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-11)+31}{2*6}=\frac{42}{12} =3+1/2 $
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