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2x^{2}-12x+2=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 2}}{2\times 2}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, -12 for b, and 2 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-\left(-12\right)±\sqrt{144-4\times 2\times 2}}{2\times 2}
Square -12.
x=\frac{-\left(-12\right)±\sqrt{144-8\times 2}}{2\times 2}
Multiply -4 times 2.
x=\frac{-\left(-12\right)±\sqrt{144-16}}{2\times 2}
Multiply -8 times 2.
x=\frac{-\left(-12\right)±\sqrt{128}}{2\times 2}
Add 144 to -16.
x=\frac{-\left(-12\right)±8\sqrt{2}}{2\times 2}
Take the square root of 128.
x=\frac{12±8\sqrt{2}}{2\times 2}
The opposite of -12 is 12.
x=\frac{12±8\sqrt{2}}{4}
Multiply 2 times 2.
x=\frac{8\sqrt{2}+12}{4}
Now solve the equation x=\frac{12±8\sqrt{2}}{4} when ± is plus. Add 12 to 8\sqrt{2}.
x=2\sqrt{2}+3
Divide 12+8\sqrt{2} by 4.
x=\frac{12-8\sqrt{2}}{4}
Now solve the equation x=\frac{12±8\sqrt{2}}{4} when ± is minus. Subtract 8\sqrt{2} from 12.
x=3-2\sqrt{2}
Divide 12-8\sqrt{2} by 4.
x=2\sqrt{2}+3 x=3-2\sqrt{2}
The equation is now solved.
2x^{2}-12x+2=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+2-2=-2
Subtract 2 from both sides of the equation.
2x^{2}-12x=-2
Subtracting 2 from itself leaves 0.
\frac{2x^{2}-12x}{2}=-\frac{2}{2}
Divide both sides by 2.
x^{2}+\left(-\frac{12}{2}\right)x=-\frac{2}{2}
Dividing by 2 undoes the multiplication by 2.
x^{2}-6x=-\frac{2}{2}
Divide -12 by 2.
x^{2}-6x=-1
Divide -2 by 2.
x^{2}-6x+\left(-3\right)^{2}=-1+\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=-1+9
Square -3.
x^{2}-6x+9=8
Add -1 to 9.
\left(x-3\right)^{2}=8
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{8}
Take the square root of both sides of the equation.
x-3=2\sqrt{2} x-3=-2\sqrt{2}
Simplify.
x=2\sqrt{2}+3 x=3-2\sqrt{2}
Add 3 to both sides of the equation.
x ^ 2 -6x +1 = 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 = 1
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) = 1
To solve for unknown quantity u, substitute these in the product equation rs = 1
9 - u^2 = 1
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = 1-9 = -8
Simplify the expression by subtracting 9 on both sides
u^2 = 8 u = \pm\sqrt{8} = \pm \sqrt{8}
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{8} = 0.172 s = 3 + \sqrt{8} = 5.828
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.