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3x^{2}+5x+6=0
All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.
x=\frac{-5±\sqrt{5^{2}-4\times 3\times 6}}{2\times 3}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, 5 for b, and 6 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-5±\sqrt{25-4\times 3\times 6}}{2\times 3}
Square 5.
x=\frac{-5±\sqrt{25-12\times 6}}{2\times 3}
Multiply -4 times 3.
x=\frac{-5±\sqrt{25-72}}{2\times 3}
Multiply -12 times 6.
x=\frac{-5±\sqrt{-47}}{2\times 3}
Add 25 to -72.
x=\frac{-5±\sqrt{47}i}{2\times 3}
Take the square root of -47.
x=\frac{-5±\sqrt{47}i}{6}
Multiply 2 times 3.
x=\frac{-5+\sqrt{47}i}{6}
Now solve the equation x=\frac{-5±\sqrt{47}i}{6} when ± is plus. Add -5 to i\sqrt{47}.
x=\frac{-\sqrt{47}i-5}{6}
Now solve the equation x=\frac{-5±\sqrt{47}i}{6} when ± is minus. Subtract i\sqrt{47} from -5.
x=\frac{-5+\sqrt{47}i}{6} x=\frac{-\sqrt{47}i-5}{6}
The equation is now solved.
3x^{2}+5x+6=0
Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.
3x^{2}+5x+6-6=-6
Subtract 6 from both sides of the equation.
3x^{2}+5x=-6
Subtracting 6 from itself leaves 0.
\frac{3x^{2}+5x}{3}=-\frac{6}{3}
Divide both sides by 3.
x^{2}+\frac{5}{3}x=-\frac{6}{3}
Dividing by 3 undoes the multiplication by 3.
x^{2}+\frac{5}{3}x=-2
Divide -6 by 3.
x^{2}+\frac{5}{3}x+\left(\frac{5}{6}\right)^{2}=-2+\left(\frac{5}{6}\right)^{2}
Divide \frac{5}{3}, the coefficient of the x term, by 2 to get \frac{5}{6}. Then add the square of \frac{5}{6} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}+\frac{5}{3}x+\frac{25}{36}=-2+\frac{25}{36}
Square \frac{5}{6} by squaring both the numerator and the denominator of the fraction.
x^{2}+\frac{5}{3}x+\frac{25}{36}=-\frac{47}{36}
Add -2 to \frac{25}{36}.
\left(x+\frac{5}{6}\right)^{2}=-\frac{47}{36}
Factor x^{2}+\frac{5}{3}x+\frac{25}{36}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.
\sqrt{\left(x+\frac{5}{6}\right)^{2}}=\sqrt{-\frac{47}{36}}
Take the square root of both sides of the equation.
x+\frac{5}{6}=\frac{\sqrt{47}i}{6} x+\frac{5}{6}=-\frac{\sqrt{47}i}{6}
Simplify.
x=\frac{-5+\sqrt{47}i}{6} x=\frac{-\sqrt{47}i-5}{6}
Subtract \frac{5}{6} from both sides of the equation.