Solve for x
x = \frac{\sqrt{7} + 5}{6} \approx 1.274291885
x=\frac{5-\sqrt{7}}{6}\approx 0.392374781
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6x^{2}-10x+3=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(-10\right)±\sqrt{\left(-10\right)^{2}-4\times 6\times 3}}{2\times 6}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 6 for a, -10 for b, and 3 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-\left(-10\right)±\sqrt{100-4\times 6\times 3}}{2\times 6}
Square -10.
x=\frac{-\left(-10\right)±\sqrt{100-24\times 3}}{2\times 6}
Multiply -4 times 6.
x=\frac{-\left(-10\right)±\sqrt{100-72}}{2\times 6}
Multiply -24 times 3.
x=\frac{-\left(-10\right)±\sqrt{28}}{2\times 6}
Add 100 to -72.
x=\frac{-\left(-10\right)±2\sqrt{7}}{2\times 6}
Take the square root of 28.
x=\frac{10±2\sqrt{7}}{2\times 6}
The opposite of -10 is 10.
x=\frac{10±2\sqrt{7}}{12}
Multiply 2 times 6.
x=\frac{2\sqrt{7}+10}{12}
Now solve the equation x=\frac{10±2\sqrt{7}}{12} when ± is plus. Add 10 to 2\sqrt{7}.
x=\frac{\sqrt{7}+5}{6}
Divide 10+2\sqrt{7} by 12.
x=\frac{10-2\sqrt{7}}{12}
Now solve the equation x=\frac{10±2\sqrt{7}}{12} when ± is minus. Subtract 2\sqrt{7} from 10.
x=\frac{5-\sqrt{7}}{6}
Divide 10-2\sqrt{7} by 12.
x=\frac{\sqrt{7}+5}{6} x=\frac{5-\sqrt{7}}{6}
The equation is now solved.
6x^{2}-10x+3=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}-10x+3-3=-3
Subtract 3 from both sides of the equation.
6x^{2}-10x=-3
Subtracting 3 from itself leaves 0.
\frac{6x^{2}-10x}{6}=-\frac{3}{6}
Divide both sides by 6.
x^{2}+\left(-\frac{10}{6}\right)x=-\frac{3}{6}
Dividing by 6 undoes the multiplication by 6.
x^{2}-\frac{5}{3}x=-\frac{3}{6}
Reduce the fraction \frac{-10}{6} to lowest terms by extracting and canceling out 2.
x^{2}-\frac{5}{3}x=-\frac{1}{2}
Reduce the fraction \frac{-3}{6} to lowest terms by extracting and canceling out 3.
x^{2}-\frac{5}{3}x+\left(-\frac{5}{6}\right)^{2}=-\frac{1}{2}+\left(-\frac{5}{6}\right)^{2}
Divide -\frac{5}{3}, the coefficient of the x term, by 2 to get -\frac{5}{6}. Then add the square of -\frac{5}{6} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}-\frac{5}{3}x+\frac{25}{36}=-\frac{1}{2}+\frac{25}{36}
Square -\frac{5}{6} by squaring both the numerator and the denominator of the fraction.
x^{2}-\frac{5}{3}x+\frac{25}{36}=\frac{7}{36}
Add -\frac{1}{2} to \frac{25}{36} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
\left(x-\frac{5}{6}\right)^{2}=\frac{7}{36}
Factor x^{2}-\frac{5}{3}x+\frac{25}{36}. 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}{6}\right)^{2}}=\sqrt{\frac{7}{36}}
Take the square root of both sides of the equation.
x-\frac{5}{6}=\frac{\sqrt{7}}{6} x-\frac{5}{6}=-\frac{\sqrt{7}}{6}
Simplify.
x=\frac{\sqrt{7}+5}{6} x=\frac{5-\sqrt{7}}{6}
Add \frac{5}{6} to both sides of the equation.
x ^ 2 -\frac{5}{3}x +\frac{1}{2} = 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}{3} rs = \frac{1}{2}
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}{6} - u s = \frac{5}{6} + u
Two numbers r and s sum up to \frac{5}{3} exactly when the average of the two numbers is \frac{1}{2}*\frac{5}{3} = \frac{5}{6}. 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}{6} - u) (\frac{5}{6} + u) = \frac{1}{2}
To solve for unknown quantity u, substitute these in the product equation rs = \frac{1}{2}
\frac{25}{36} - u^2 = \frac{1}{2}
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = \frac{1}{2}-\frac{25}{36} = -\frac{7}{36}
Simplify the expression by subtracting \frac{25}{36} on both sides
u^2 = \frac{7}{36} u = \pm\sqrt{\frac{7}{36}} = \pm \frac{\sqrt{7}}{6}
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}{6} - \frac{\sqrt{7}}{6} = 0.392 s = \frac{5}{6} + \frac{\sqrt{7}}{6} = 1.274
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.
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