Question

5t²+20t−(t+4)=0

x1=
x2=

Answer 1

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

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

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

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

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

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