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3x^{2}-2x+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{-\left(-2\right)±\sqrt{\left(-2\right)^{2}-4\times 3\times 6}}{2\times 3}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, -2 for b, and 6 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-\left(-2\right)±\sqrt{4-4\times 3\times 6}}{2\times 3}
Square -2.
x=\frac{-\left(-2\right)±\sqrt{4-12\times 6}}{2\times 3}
Multiply -4 times 3.
x=\frac{-\left(-2\right)±\sqrt{4-72}}{2\times 3}
Multiply -12 times 6.
x=\frac{-\left(-2\right)±\sqrt{-68}}{2\times 3}
Add 4 to -72.
x=\frac{-\left(-2\right)±2\sqrt{17}i}{2\times 3}
Take the square root of -68.
x=\frac{2±2\sqrt{17}i}{2\times 3}
The opposite of -2 is 2.
x=\frac{2±2\sqrt{17}i}{6}
Multiply 2 times 3.
x=\frac{2+2\sqrt{17}i}{6}
Now solve the equation x=\frac{2±2\sqrt{17}i}{6} when ± is plus. Add 2 to 2i\sqrt{17}.
x=\frac{1+\sqrt{17}i}{3}
Divide 2+2i\sqrt{17} by 6.
x=\frac{-2\sqrt{17}i+2}{6}
Now solve the equation x=\frac{2±2\sqrt{17}i}{6} when ± is minus. Subtract 2i\sqrt{17} from 2.
x=\frac{-\sqrt{17}i+1}{3}
Divide 2-2i\sqrt{17} by 6.
x=\frac{1+\sqrt{17}i}{3} x=\frac{-\sqrt{17}i+1}{3}
The equation is now solved.
3x^{2}-2x+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}-2x+6-6=-6
Subtract 6 from both sides of the equation.
3x^{2}-2x=-6
Subtracting 6 from itself leaves 0.
\frac{3x^{2}-2x}{3}=-\frac{6}{3}
Divide both sides by 3.
x^{2}-\frac{2}{3}x=-\frac{6}{3}
Dividing by 3 undoes the multiplication by 3.
x^{2}-\frac{2}{3}x=-2
Divide -6 by 3.
x^{2}-\frac{2}{3}x+\left(-\frac{1}{3}\right)^{2}=-2+\left(-\frac{1}{3}\right)^{2}
Divide -\frac{2}{3}, the coefficient of the x term, by 2 to get -\frac{1}{3}. Then add the square of -\frac{1}{3} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}-\frac{2}{3}x+\frac{1}{9}=-2+\frac{1}{9}
Square -\frac{1}{3} by squaring both the numerator and the denominator of the fraction.
x^{2}-\frac{2}{3}x+\frac{1}{9}=-\frac{17}{9}
Add -2 to \frac{1}{9}.
\left(x-\frac{1}{3}\right)^{2}=-\frac{17}{9}
Factor x^{2}-\frac{2}{3}x+\frac{1}{9}. 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{1}{3}\right)^{2}}=\sqrt{-\frac{17}{9}}
Take the square root of both sides of the equation.
x-\frac{1}{3}=\frac{\sqrt{17}i}{3} x-\frac{1}{3}=-\frac{\sqrt{17}i}{3}
Simplify.
x=\frac{1+\sqrt{17}i}{3} x=\frac{-\sqrt{17}i+1}{3}
Add \frac{1}{3} to both sides of the equation.
x ^ 2 -\frac{2}{3}x +2 = 0
Quadratic equations such as this one can be solved by a new direct factoring method that does not require guess work. To use the direct factoring method, the equation must be in the form x^2+Bx+C=0.This is achieved by dividing both sides of the equation by 3
r + s = \frac{2}{3} rs = 2
Let r and s be the factors for the quadratic equation such that x^2+Bx+C=(x−r)(x−s) where sum of factors (r+s)=−B and the product of factors rs = C
r = \frac{1}{3} - u s = \frac{1}{3} + u
Two numbers r and s sum up to \frac{2}{3} exactly when the average of the two numbers is \frac{1}{2}*\frac{2}{3} = \frac{1}{3}. You can also see that the midpoint of r and s corresponds to the axis of symmetry of the parabola represented by the quadratic equation y=x^2+Bx+C. The values of r and s are equidistant from the center by an unknown quantity u. Express r and s with respect to variable u. <div style='padding: 8px'><img src='https://opalmath.azureedge.net/customsolver/quadraticgraph.png' style='width: 100%;max-width: 700px' /></div>
(\frac{1}{3} - u) (\frac{1}{3} + u) = 2
To solve for unknown quantity u, substitute these in the product equation rs = 2
\frac{1}{9} - u^2 = 2
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = 2-\frac{1}{9} = \frac{17}{9}
Simplify the expression by subtracting \frac{1}{9} on both sides
u^2 = -\frac{17}{9} u = \pm\sqrt{-\frac{17}{9}} = \pm \frac{\sqrt{17}}{3}i
Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u
r =\frac{1}{3} - \frac{\sqrt{17}}{3}i = 0.333 - 1.374i s = \frac{1}{3} + \frac{\sqrt{17}}{3}i = 0.333 + 1.374i
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.