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14x^{2}-23x+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(-23\right)±\sqrt{\left(-23\right)^{2}-4\times 14\times 6}}{2\times 14}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 14 for a, -23 for b, and 6 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-\left(-23\right)±\sqrt{529-4\times 14\times 6}}{2\times 14}
Square -23.
x=\frac{-\left(-23\right)±\sqrt{529-56\times 6}}{2\times 14}
Multiply -4 times 14.
x=\frac{-\left(-23\right)±\sqrt{529-336}}{2\times 14}
Multiply -56 times 6.
x=\frac{-\left(-23\right)±\sqrt{193}}{2\times 14}
Add 529 to -336.
x=\frac{23±\sqrt{193}}{2\times 14}
The opposite of -23 is 23.
x=\frac{23±\sqrt{193}}{28}
Multiply 2 times 14.
x=\frac{\sqrt{193}+23}{28}
Now solve the equation x=\frac{23±\sqrt{193}}{28} when ± is plus. Add 23 to \sqrt{193}.
x=\frac{23-\sqrt{193}}{28}
Now solve the equation x=\frac{23±\sqrt{193}}{28} when ± is minus. Subtract \sqrt{193} from 23.
x=\frac{\sqrt{193}+23}{28} x=\frac{23-\sqrt{193}}{28}
The equation is now solved.
14x^{2}-23x+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.
14x^{2}-23x+6-6=-6
Subtract 6 from both sides of the equation.
14x^{2}-23x=-6
Subtracting 6 from itself leaves 0.
\frac{14x^{2}-23x}{14}=-\frac{6}{14}
Divide both sides by 14.
x^{2}-\frac{23}{14}x=-\frac{6}{14}
Dividing by 14 undoes the multiplication by 14.
x^{2}-\frac{23}{14}x=-\frac{3}{7}
Reduce the fraction \frac{-6}{14} to lowest terms by extracting and canceling out 2.
x^{2}-\frac{23}{14}x+\left(-\frac{23}{28}\right)^{2}=-\frac{3}{7}+\left(-\frac{23}{28}\right)^{2}
Divide -\frac{23}{14}, the coefficient of the x term, by 2 to get -\frac{23}{28}. Then add the square of -\frac{23}{28} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}-\frac{23}{14}x+\frac{529}{784}=-\frac{3}{7}+\frac{529}{784}
Square -\frac{23}{28} by squaring both the numerator and the denominator of the fraction.
x^{2}-\frac{23}{14}x+\frac{529}{784}=\frac{193}{784}
Add -\frac{3}{7} to \frac{529}{784} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
\left(x-\frac{23}{28}\right)^{2}=\frac{193}{784}
Factor x^{2}-\frac{23}{14}x+\frac{529}{784}. 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{23}{28}\right)^{2}}=\sqrt{\frac{193}{784}}
Take the square root of both sides of the equation.
x-\frac{23}{28}=\frac{\sqrt{193}}{28} x-\frac{23}{28}=-\frac{\sqrt{193}}{28}
Simplify.
x=\frac{\sqrt{193}+23}{28} x=\frac{23-\sqrt{193}}{28}
Add \frac{23}{28} to both sides of the equation.
x ^ 2 -\frac{23}{14}x +\frac{3}{7} = 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 14
r + s = \frac{23}{14} rs = \frac{3}{7}
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{23}{28} - u s = \frac{23}{28} + u
Two numbers r and s sum up to \frac{23}{14} exactly when the average of the two numbers is \frac{1}{2}*\frac{23}{14} = \frac{23}{28}. 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{23}{28} - u) (\frac{23}{28} + u) = \frac{3}{7}
To solve for unknown quantity u, substitute these in the product equation rs = \frac{3}{7}
\frac{529}{784} - u^2 = \frac{3}{7}
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = \frac{3}{7}-\frac{529}{784} = -\frac{193}{784}
Simplify the expression by subtracting \frac{529}{784} on both sides
u^2 = \frac{193}{784} u = \pm\sqrt{\frac{193}{784}} = \pm \frac{\sqrt{193}}{28}
Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u
r =\frac{23}{28} - \frac{\sqrt{193}}{28} = 0.325 s = \frac{23}{28} + \frac{\sqrt{193}}{28} = 1.318
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.