Question

solve root3x^2+root2x+2root3​

Answer 1

Answer:

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Step-by-step explanation:

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Answer 2

$\huge\red{\mid{\underline{\overline{\textbf{SOLuTIoN}}}\mid}}$

a = $\sqrt{3}$

b = 2$\sqrt{2}$

c = 2$\sqrt{3}$

☆ Formula☆

$\huge{\boxed{\blue{{b}^{2} - 4ac}}}$

$D = {(2 \sqrt{2}) }^{2} - 4 \times \: \sqrt{3} \times ( - 2 \sqrt{3} ) \: \\ D = 4 \times 2 \: - \:4 \sqrt{3} \: \times \: 2 \sqrt{3} \\ D = \: 8 \: + \: 8 \: \times \: 3\\ D = 8 \: + \: 24 \\ D = 32$

☆ Formula☆

$\huge{\boxed{\orange {\frac{ - b \frac{ + }{ } \sqrt{D} }{ 2a}}}}$

$x = \frac{ - 2 \sqrt{2} \frac{ + }{} \sqrt{32} }{2×3} \\ x = \frac{2 \sqrt{2 } \frac{ + }{}4 \sqrt{2} }{2 \sqrt{3} } \\ x = \frac{2 \sqrt{2} + 4 \sqrt{2} }{2 \sqrt{3} } \\ x = \frac{6 \sqrt{2} }{2 \sqrt{3} } \\ \ x = \frac{3 \sqrt{2} }{ \sqrt{3} } \\ \\ \: \: \: \: \: \: \: or \\ \\ x = \frac{2 \sqrt{2 } - 4 \sqrt{2} }{2 \sqrt{3} } \\x = \frac{ - 2 \sqrt{2} }{2 \sqrt{3} } \\ x = \frac{ - \sqrt{2} }{ \sqrt{3} }$

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