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-3t^2-4t+4=0
a = -3; b = -4; c = +4;
Δ = b2-4ac
Δ = -42-4·(-3)·4
Δ = 64
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{64}=8$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-8}{2*-3}=\frac{-4}{-6} =2/3 $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+8}{2*-3}=\frac{12}{-6} =-2 $
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