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2x^{2}-4x+12=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(-4\right)±\sqrt{\left(-4\right)^{2}-4\times 2\times 12}}{2\times 2}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, -4 for b, and 12 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-\left(-4\right)±\sqrt{16-4\times 2\times 12}}{2\times 2}
Square -4.
x=\frac{-\left(-4\right)±\sqrt{16-8\times 12}}{2\times 2}
Multiply -4 times 2.
x=\frac{-\left(-4\right)±\sqrt{16-96}}{2\times 2}
Multiply -8 times 12.
x=\frac{-\left(-4\right)±\sqrt{-80}}{2\times 2}
Add 16 to -96.
x=\frac{-\left(-4\right)±4\sqrt{5}i}{2\times 2}
Take the square root of -80.
x=\frac{4±4\sqrt{5}i}{2\times 2}
The opposite of -4 is 4.
x=\frac{4±4\sqrt{5}i}{4}
Multiply 2 times 2.
x=\frac{4+4\sqrt{5}i}{4}
Now solve the equation x=\frac{4±4\sqrt{5}i}{4} when ± is plus. Add 4 to 4i\sqrt{5}.
x=1+\sqrt{5}i
Divide 4+4i\sqrt{5} by 4.
x=\frac{-4\sqrt{5}i+4}{4}
Now solve the equation x=\frac{4±4\sqrt{5}i}{4} when ± is minus. Subtract 4i\sqrt{5} from 4.
x=-\sqrt{5}i+1
Divide 4-4i\sqrt{5} by 4.
x=1+\sqrt{5}i x=-\sqrt{5}i+1
The equation is now solved.
2x^{2}-4x+12=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}-4x+12-12=-12
Subtract 12 from both sides of the equation.
2x^{2}-4x=-12
Subtracting 12 from itself leaves 0.
\frac{2x^{2}-4x}{2}=-\frac{12}{2}
Divide both sides by 2.
x^{2}+\left(-\frac{4}{2}\right)x=-\frac{12}{2}
Dividing by 2 undoes the multiplication by 2.
x^{2}-2x=-\frac{12}{2}
Divide -4 by 2.
x^{2}-2x=-6
Divide -12 by 2.
x^{2}-2x+1=-6+1
Divide -2, the coefficient of the x term, by 2 to get -1. Then add the square of -1 to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}-2x+1=-5
Add -6 to 1.
\left(x-1\right)^{2}=-5
Factor x^{2}-2x+1. 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-1\right)^{2}}=\sqrt{-5}
Take the square root of both sides of the equation.
x-1=\sqrt{5}i x-1=-\sqrt{5}i
Simplify.
x=1+\sqrt{5}i x=-\sqrt{5}i+1
Add 1 to both sides of the equation.
x ^ 2 -2x +6 = 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 = 2 rs = 6
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 = 1 - u s = 1 + u
Two numbers r and s sum up to 2 exactly when the average of the two numbers is \frac{1}{2}*2 = 1. 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>
(1 - u) (1 + u) = 6
To solve for unknown quantity u, substitute these in the product equation rs = 6
1 - u^2 = 6
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = 6-1 = 5
Simplify the expression by subtracting 1 on both sides
u^2 = -5 u = \pm\sqrt{-5} = \pm \sqrt{5}i
Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u
r =1 - \sqrt{5}i s = 1 + \sqrt{5}i
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.