Question

2(3x+1)²=0 with compete solution ​

Answer 1

$\huge\sf\fbox\red{Question:-}$

${2(3x + 1)}^{2} = 0$

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$18x ^{2} + 6x + 1$

$\huge\sf\fbox\red{Solution:-}$

${( a + b)}^{2} = ({a})^{2} + 2ab + ( {b})^{2}$

$2 \times (3 {x}^{2} ) + 2 \times 3x \times 1 + ( {1})^{2}$

$2 \times {9x}^{2} + 2 \times 3x \times 1 + 1$

$18x ^{2} + 6x + 1$

hope it helps