4. Solve `t` `-3 = 2t +1/2(-10)t^2`
4. Solve t -3 = 2t +1/2(-10)t^2
Hint
`t = (-b ± sqrt(b^2-4ac))/(2a)`
Answer
` t_1 = - 0.6, t_2 = 1`
Show Steps
1) Simplify equation: `=> -5t^2 + 2t +3 = 0` 2) Use quadratic formula: `t = (-b ± sqrt(b^2-4ac))/(2a)` where `a = -5` `b =2 ` `c =3` `=> t = (-2 ± sqrt((2)^2-4(-5)(3)))/(2(-5))` `=> t = (-2±8)/-10` `=> t_1 = - 0.6, t_2 = 1`