$7(x-3)^2-4(x-3)-3\\\\substitute\ t=x-3,\ then\ we\ have\\\\7t^2-4t-3\\\\a=7;\ b=-4;\ c=-3\\\\\Delta=b^2-4ac\\\\\Delta=(-4)^2-4\cdot7\cdot(-3)=16+84=100\\\\t_1=\dfrac{-b-\sqrt\Delta}{2a};\ t_2=\dfrac{-b+\sqrt\Delta}{2a}\\\\\sqrt\Delta=\sqrt{100}=10\\\\t_1=\dfrac{-(-4)-10}{2\cdot7}=\dfrac{4-10}{14}=\dfrac{-6}{14}=-\dfrac{3}{7}\\\\t_2=\dfrac{-(-4)+10}{2\cdot7}=\dfrac{4+10}{14}=\dfrac{14}{14}=1\\\\7t^2-4t-3=7\left(t+\dfrac{3}{7}\right)(t-1)=(7t+3)(t-1)\\\\=(7(x-3)+3)(x-3-1)=(7x-21+3)(x-4)\\\\=\boxed{(7x-18)(x-4)}$

7 0

4 years ago

On a trip you had to change your money from dollars to British pounds. You got 560 pounds for 800 dollars. What is a unit rate t

Oksanka [162]

If we got 560 pounds for 800 dollars and x pounds for 1 dollar, this means that x = 560*1/800 <=> x = 56/80. Performing the calulation we get that x = 0.7

6 0

4 years ago

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