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What are the approximate solutions of 4x2 + 3 = −12x to the nearest hundredth? A. x ≈ −3.23 and x ≈ 0.23 B.x ≈ −2.72 and x ≈ −0.28 C.x ≈ 0.28 and x ≈ 2.72 D.x ≈ −0.23 and x ≈ 3.23
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\(\bf 4x^2 + 3 = -12x\implies 4x^2+3+12x=0\implies {\color{blue}{ 4}}x^2{\color{red}{ +12}}x{\color{green}{ +3}}=0 \\ \quad \\ \textit{quadratic formula}\\ \qquad \qquad x= \cfrac{ - {\color{red}{ b}} \pm \sqrt { {\color{red}{ b}}^2 -4{\color{blue}{ a}}{\color{green}{ c}}}}{2{\color{blue}{ a}}}\)
so b
a= 4 b=12 c=3 Put those numbers into the quadratic formula and you'll have your answer.
Yep as wolf said, Use quadratic formula with a=4, b=12, c=3.
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