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4t^2-3t=27
We move all terms to the left:
4t^2-3t-(27)=0
a = 4; b = -3; c = -27;
Δ = b2-4ac
Δ = -32-4·4·(-27)
Δ = 441
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{441}=21$$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-21}{2*4}=\frac{-18}{8} =-2+1/4 $$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+21}{2*4}=\frac{24}{8} =3 $
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