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2x^{2}-12x+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(-12\right)±\sqrt{\left(-12\right)^{2}-4\times 2\times 6}}{2\times 2}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, -12 for b, and 6 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-\left(-12\right)±\sqrt{144-4\times 2\times 6}}{2\times 2}
Square -12.
x=\frac{-\left(-12\right)±\sqrt{144-8\times 6}}{2\times 2}
Multiply -4 times 2.
x=\frac{-\left(-12\right)±\sqrt{144-48}}{2\times 2}
Multiply -8 times 6.
x=\frac{-\left(-12\right)±\sqrt{96}}{2\times 2}
Add 144 to -48.
x=\frac{-\left(-12\right)±4\sqrt{6}}{2\times 2}
Take the square root of 96.
x=\frac{12±4\sqrt{6}}{2\times 2}
The opposite of -12 is 12.
x=\frac{12±4\sqrt{6}}{4}
Multiply 2 times 2.
x=\frac{4\sqrt{6}+12}{4}
Now solve the equation x=\frac{12±4\sqrt{6}}{4} when ± is plus. Add 12 to 4\sqrt{6}.
x=\sqrt{6}+3
Divide 12+4\sqrt{6} by 4.
x=\frac{12-4\sqrt{6}}{4}
Now solve the equation x=\frac{12±4\sqrt{6}}{4} when ± is minus. Subtract 4\sqrt{6} from 12.
x=3-\sqrt{6}
Divide 12-4\sqrt{6} by 4.
x=\sqrt{6}+3 x=3-\sqrt{6}
The equation is now solved.
2x^{2}-12x+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.
2x^{2}-12x+6-6=-6
Subtract 6 from both sides of the equation.
2x^{2}-12x=-6
Subtracting 6 from itself leaves 0.
\frac{2x^{2}-12x}{2}=-\frac{6}{2}
Divide both sides by 2.
x^{2}+\left(-\frac{12}{2}\right)x=-\frac{6}{2}
Dividing by 2 undoes the multiplication by 2.
x^{2}-6x=-\frac{6}{2}
Divide -12 by 2.
x^{2}-6x=-3
Divide -6 by 2.
x^{2}-6x+\left(-3\right)^{2}=-3+\left(-3\right)^{2}
Divide -6, the coefficient of the x term, by 2 to get -3. Then add the square of -3 to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}-6x+9=-3+9
Square -3.
x^{2}-6x+9=6
Add -3 to 9.
\left(x-3\right)^{2}=6
Factor x^{2}-6x+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-3\right)^{2}}=\sqrt{6}
Take the square root of both sides of the equation.
x-3=\sqrt{6} x-3=-\sqrt{6}
Simplify.
x=\sqrt{6}+3 x=3-\sqrt{6}
Add 3 to both sides of the equation.
x ^ 2 -6x +3 = 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 2
r + s = 6 rs = 3
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 = 3 - u s = 3 + u
Two numbers r and s sum up to 6 exactly when the average of the two numbers is \frac{1}{2}*6 = 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>
(3 - u) (3 + u) = 3
To solve for unknown quantity u, substitute these in the product equation rs = 3
9 - u^2 = 3
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = 3-9 = -6
Simplify the expression by subtracting 9 on both sides
u^2 = 6 u = \pm\sqrt{6} = \pm \sqrt{6}
Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u
r =3 - \sqrt{6} = 0.551 s = 3 + \sqrt{6} = 5.449
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.