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# Answer to Question #101342 in Mechanics | Relativity for Aman

Question #101342
Consider a equation for t below 191×2=3.97t×t+33.4t What value of t satisfies this equation
1
2020-01-20T05:24:03-0500

Given that,

@$\Rightarrow 191×2=3.97t×t+33.4t @$

@$\Rightarrow 382=3.97t^2+33.4t@$

@$\Rightarrow 3.97t^2+33.4t-382=0@$

it is the quadratic equation, so it will have 2 roots.

Now, from the quadratic equation formula,

@$\Rightarrow t=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}@$

@$\Rightarrow t=\dfrac{-33.4\pm\sqrt{(33.4)^2+4\times 3.97\times382}}{2\times 3.97}@$

@$\Rightarrow t=\dfrac{-33.4\pm\sqrt{1115.56+6066.16}}{7.94}@$

@$\Rightarrow t=\dfrac{-33.4\pm \sqrt{7181.72}}{7.94}@$

@$\Rightarrow t=\dfrac{-33.4\pm84.74}{7.94}@$

Taking "+",

@$\Rightarrow t=\dfrac{-33.4+84.74}{7.94}=6.46 @$

so,

@$t=6.5@$

@$​​ @$

Taking "-"

@$t=\dfrac{-118.14}{7.94}=-14.88@$

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