4. Solve `t`
`-3 = 2t +1/2(-10)t^2`
4. Solve t
-3 = 2t +1/2(-10)t^2 `t = (-b ± sqrt(b^2-4ac))/(2a)`
` t_1 = - 0.6, t_2 = 1`
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`
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