Question

find the zeroes of the quadratic polynomial t²+4√3t-15​

Answer 1

Step-by-step explanation:

Comparing the equation with general form, we get, a = 1 , b = 4√3, c = -15...

Applying Quadratic formula, and substituting the values.. we get,

$\frac{ - 4 \sqrt{3} + - \sqrt{ {(4 \sqrt{3}) }^{2} - 4(1)( - 15) } }{2(1)} \\ \\ = > \frac{ - 4 \sqrt{3} + - \sqrt{48 + 108} }{2}$

$= > \frac{ - 4 \sqrt{3} + - 6 \sqrt{3} }{2} \\ \\ = > -2 \sqrt{3} + 3 \sqrt{3} \: \: \: and \: \: \: - 2 \sqrt{3} - 3 \sqrt{3}$

$= > \sqrt{3} \: \: and \: \: - 5 \sqrt{3}$

Therefore, $√3$ and $- 5√3$ are the roots if the equation.....

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