![3x^2+12x=3\\\\\Rightarrow 3(x^2+4x)=3\\\\\Rightarrow x^2+4x=1\\\\\Rightarrow x^2+2\times x\times2+2^2=1+2^2\\\\\Rightarrow (x+2)^2=1+4\\\\\Rightarrow (x+2)^2=5\\\\\Rightarrow x+2=\pm\sqrt5~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~[\textup{taking square root on both sides}]\\\\\Rightarrow x=-2\pm\sqrt5.](/tpl/images/0425/5222/d383f.png)
−2 ± square root 13
−2 ± square root 7
−2 ± square root 3
Mathematics, 19.12.2019 04:31 keke2933
Solve 3x2 + 12x = 3.
−2 ± square root 13
−2 ± square root 7
−2 ± square root 3
−2 ± square root 5
Answers: 3
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You know the right answer?
Solve 3x2 + 12x = 3.
−2 ± square root 13
−2 ± square root 7
−2 ± square root 3
−2 ± square root 13
−2 ± square root 7
−2 ± square root 3
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