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JoshEast

\(3 x^{2} + 12x = 3\)
\(3 x^{2} + 12x - 3 = 0\)
Using the quadratic formula
\(x = \frac{-b (+ or -) \sqrt{b^{2 - 4ac} } }{2a}\)
\(x = \frac{-12 + \sqrt{12^{2 - 4(3)(-3)} } }{2(3)}\)   OR   \(x = \frac{-12 - \sqrt{12^{2 - 4(3)(-3)} } }{2(3)}\)
\(x = \frac{-12 + \sqrt{180} }{6}\)   OR   \(x = \frac{-12 - \sqrt{180} }{6}\)
∴ x = 0.236  and  x = -4.236 

Answer:

The solutions to the quadratic equation are: \(x=-2+\sqrt{5},\:x=-2-\sqrt{5}\)

Step-by-step explanation:

To solve the quadratic equation \(3x^2+12x=3\) you must:

Subtract 3 from both sides and simplify

\(3x^2+12x-3=3-3\\\\3x^2+12x-3=0\)

For a quadratic equation of the form \(ax^2+bx+c=0\) the solutions are

\(x_{1,\:2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\)

\(\mathrm{For\:}\quad a=3,\:b=12,\:c=-3\)

\(x=\frac{-12+\sqrt{12^2-4\cdot \:3\left(-3\right)}}{2\cdot \:3}=\frac{-12+\sqrt{180}}{2\cdot \:3}=\frac{-12+6\sqrt{5}}{6}=\frac{6\left(-2+\sqrt{5}\right)}{6}=-2+\sqrt{5}\)

\(x=\frac{-12-\sqrt{12^2-4\cdot \:3\left(-3\right)}}{2\cdot \:3}=\frac{-12-\sqrt{180}}{2\cdot \:3}=\frac{-12-6\sqrt{5}}{6}=-\frac{6\left(2+\sqrt{5}\right)}{6}=-2-\sqrt{5}\)

The solutions to the quadratic equation are:

\(x=-2+\sqrt{5},\:x=-2-\sqrt{5}\)

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