Question

Solve 3x2 + 12x = 3.

Answer 1

$3 x^{2} + 12x = 3$
$3 x^{2} + 12x - 3 = 0$
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 2

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:

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