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6x^{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{-\left(-5\right)±\sqrt{\left(-5\right)^{2}-4\times 6\times 6}}{2\times 6}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 6 for a, -5 for b, and 6 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-\left(-5\right)±\sqrt{25-4\times 6\times 6}}{2\times 6}
Square -5.
x=\frac{-\left(-5\right)±\sqrt{25-24\times 6}}{2\times 6}
Multiply -4 times 6.
x=\frac{-\left(-5\right)±\sqrt{25-144}}{2\times 6}
Multiply -24 times 6.
x=\frac{-\left(-5\right)±\sqrt{-119}}{2\times 6}
Add 25 to -144.
x=\frac{-\left(-5\right)±\sqrt{119}i}{2\times 6}
Take the square root of -119.
x=\frac{5±\sqrt{119}i}{2\times 6}
The opposite of -5 is 5.
x=\frac{5±\sqrt{119}i}{12}
Multiply 2 times 6.
x=\frac{5+\sqrt{119}i}{12}
Now solve the equation x=\frac{5±\sqrt{119}i}{12} when ± is plus. Add 5 to i\sqrt{119}.
x=\frac{-\sqrt{119}i+5}{12}
Now solve the equation x=\frac{5±\sqrt{119}i}{12} when ± is minus. Subtract i\sqrt{119} from 5.
x=\frac{5+\sqrt{119}i}{12} x=\frac{-\sqrt{119}i+5}{12}
The equation is now solved.
6x^{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.
6x^{2}-5x+6-6=-6
Subtract 6 from both sides of the equation.
6x^{2}-5x=-6
Subtracting 6 from itself leaves 0.
\frac{6x^{2}-5x}{6}=-\frac{6}{6}
Divide both sides by 6.
x^{2}-\frac{5}{6}x=-\frac{6}{6}
Dividing by 6 undoes the multiplication by 6.
x^{2}-\frac{5}{6}x=-1
Divide -6 by 6.
x^{2}-\frac{5}{6}x+\left(-\frac{5}{12}\right)^{2}=-1+\left(-\frac{5}{12}\right)^{2}
Divide -\frac{5}{6}, the coefficient of the x term, by 2 to get -\frac{5}{12}. Then add the square of -\frac{5}{12} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}-\frac{5}{6}x+\frac{25}{144}=-1+\frac{25}{144}
Square -\frac{5}{12} by squaring both the numerator and the denominator of the fraction.
x^{2}-\frac{5}{6}x+\frac{25}{144}=-\frac{119}{144}
Add -1 to \frac{25}{144}.
\left(x-\frac{5}{12}\right)^{2}=-\frac{119}{144}
Factor x^{2}-\frac{5}{6}x+\frac{25}{144}. 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}{12}\right)^{2}}=\sqrt{-\frac{119}{144}}
Take the square root of both sides of the equation.
x-\frac{5}{12}=\frac{\sqrt{119}i}{12} x-\frac{5}{12}=-\frac{\sqrt{119}i}{12}
Simplify.
x=\frac{5+\sqrt{119}i}{12} x=\frac{-\sqrt{119}i+5}{12}
Add \frac{5}{12} to both sides of the equation.
x ^ 2 -\frac{5}{6}x +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 6
r + s = \frac{5}{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 = \frac{5}{12} - u s = \frac{5}{12} + u
Two numbers r and s sum up to \frac{5}{6} exactly when the average of the two numbers is \frac{1}{2}*\frac{5}{6} = \frac{5}{12}. 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{5}{12} - u) (\frac{5}{12} + u) = 1
To solve for unknown quantity u, substitute these in the product equation rs = 1
\frac{25}{144} - u^2 = 1
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = 1-\frac{25}{144} = \frac{119}{144}
Simplify the expression by subtracting \frac{25}{144} on both sides
u^2 = -\frac{119}{144} u = \pm\sqrt{-\frac{119}{144}} = \pm \frac{\sqrt{119}}{12}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{5}{12} - \frac{\sqrt{119}}{12}i = 0.417 - 0.909i s = \frac{5}{12} + \frac{\sqrt{119}}{12}i = 0.417 + 0.909i
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.