Question

5t²+20t−(t+4)=0

x1=
x2=

Answer 1

$\displaystyle \tt 5t^{2}+20t-(t+4)=0$

$\displaystyle \tt 5t^{2}+20t-t-4=0$

$\displaystyle \tt 5t^2+19t-4=0$

$\displaystyle \tt D=b^{2}-4ac=19^2-4*5*(-4)=441$

$\displaystyle \tt t=\frac{-b-\sqrt{D} }{2a}=\frac{-19-21}{2*5}=-\frac{40}{10}=\bold\bold-\bold4$

$\displaystyle \tt t=\frac{-b+\sqrt{D} }{2a}=\frac{-19+21}{2*5}=\frac{2}{10}=\frac{\bold1}{\bold5} \bold$