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16t^2+14t-15=0
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do you know quadratic formula ?
You have to use ax^2+bx+c=0
Compare your quadratic equation with \(at^2+bt+c=0\) find \[a=...?\\b=...?\\c=...?\\\] \[ \\ \sqrt{b^2-4ac}=...?\] then the two roots of t are: \(\huge{t_{1,2}=\frac{-b \pm \sqrt{b^2-4ac}}{2a}}\)
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first step is to find a=...? b= ...? c=... ?
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a= 1 b= 4 c=3
how did you get those? is your equation, \(t^2+4t+3-0 \) ??
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