r^2 +4t-12=0
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r^2 +4t-12=0 ans the question
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Step-by-step explanation:
idk whether to solve it or factorize it, so anyways here you go with both
1) factorizing
[tex]t^{2} + 4t - 12 = 0\\\\t^{2} + 6t -2t - 12 = 0\\t(t+6) - 2 (t+6) = 0\\(t+6)(t-2) = 0\\[/tex]
2) to solve for t we know
[tex]\frac{ -b ± \sqrt{b^{2} - 4ac} }{2a}\\[/tex]
placing the values we get
[tex]\frac{-4 ± \sqrt{16+48} }{2}\\ = \frac{-4 + \sqrt{16+48} }{2}\\ , \frac{-4 - \sqrt{16+48} }{2}\\= \frac{-4+8}{2} , \frac{-4-8}{2}\\ = \frac{4}{2} , \frac{-12}{2}\\= 2 , -6\\t = 2,-6[/tex]
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