Solve for x
x=\frac{\sqrt{55}}{6}+\frac{8}{3}\approx 3.902699748
x=-\frac{\sqrt{55}}{6}+\frac{8}{3}\approx 1.430633585
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12x^{2}-64x+67=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(-64\right)±\sqrt{\left(-64\right)^{2}-4\times 12\times 67}}{2\times 12}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 12 for a, -64 for b, and 67 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-\left(-64\right)±\sqrt{4096-4\times 12\times 67}}{2\times 12}
Square -64.
x=\frac{-\left(-64\right)±\sqrt{4096-48\times 67}}{2\times 12}
Multiply -4 times 12.
x=\frac{-\left(-64\right)±\sqrt{4096-3216}}{2\times 12}
Multiply -48 times 67.
x=\frac{-\left(-64\right)±\sqrt{880}}{2\times 12}
Add 4096 to -3216.
x=\frac{-\left(-64\right)±4\sqrt{55}}{2\times 12}
Take the square root of 880.
x=\frac{64±4\sqrt{55}}{2\times 12}
The opposite of -64 is 64.
x=\frac{64±4\sqrt{55}}{24}
Multiply 2 times 12.
x=\frac{4\sqrt{55}+64}{24}
Now solve the equation x=\frac{64±4\sqrt{55}}{24} when ± is plus. Add 64 to 4\sqrt{55}.
x=\frac{\sqrt{55}}{6}+\frac{8}{3}
Divide 64+4\sqrt{55} by 24.
x=\frac{64-4\sqrt{55}}{24}
Now solve the equation x=\frac{64±4\sqrt{55}}{24} when ± is minus. Subtract 4\sqrt{55} from 64.
x=-\frac{\sqrt{55}}{6}+\frac{8}{3}
Divide 64-4\sqrt{55} by 24.
x=\frac{\sqrt{55}}{6}+\frac{8}{3} x=-\frac{\sqrt{55}}{6}+\frac{8}{3}
The equation is now solved.
12x^{2}-64x+67=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.
12x^{2}-64x+67-67=-67
Subtract 67 from both sides of the equation.
12x^{2}-64x=-67
Subtracting 67 from itself leaves 0.
\frac{12x^{2}-64x}{12}=-\frac{67}{12}
Divide both sides by 12.
x^{2}+\left(-\frac{64}{12}\right)x=-\frac{67}{12}
Dividing by 12 undoes the multiplication by 12.
x^{2}-\frac{16}{3}x=-\frac{67}{12}
Reduce the fraction \frac{-64}{12} to lowest terms by extracting and canceling out 4.
x^{2}-\frac{16}{3}x+\left(-\frac{8}{3}\right)^{2}=-\frac{67}{12}+\left(-\frac{8}{3}\right)^{2}
Divide -\frac{16}{3}, the coefficient of the x term, by 2 to get -\frac{8}{3}. Then add the square of -\frac{8}{3} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}-\frac{16}{3}x+\frac{64}{9}=-\frac{67}{12}+\frac{64}{9}
Square -\frac{8}{3} by squaring both the numerator and the denominator of the fraction.
x^{2}-\frac{16}{3}x+\frac{64}{9}=\frac{55}{36}
Add -\frac{67}{12} to \frac{64}{9} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
\left(x-\frac{8}{3}\right)^{2}=\frac{55}{36}
Factor x^{2}-\frac{16}{3}x+\frac{64}{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-\frac{8}{3}\right)^{2}}=\sqrt{\frac{55}{36}}
Take the square root of both sides of the equation.
x-\frac{8}{3}=\frac{\sqrt{55}}{6} x-\frac{8}{3}=-\frac{\sqrt{55}}{6}
Simplify.
x=\frac{\sqrt{55}}{6}+\frac{8}{3} x=-\frac{\sqrt{55}}{6}+\frac{8}{3}
Add \frac{8}{3} to both sides of the equation.
x ^ 2 -\frac{16}{3}x +\frac{67}{12} = 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 12
r + s = \frac{16}{3} rs = \frac{67}{12}
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{8}{3} - u s = \frac{8}{3} + u
Two numbers r and s sum up to \frac{16}{3} exactly when the average of the two numbers is \frac{1}{2}*\frac{16}{3} = \frac{8}{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>
(\frac{8}{3} - u) (\frac{8}{3} + u) = \frac{67}{12}
To solve for unknown quantity u, substitute these in the product equation rs = \frac{67}{12}
\frac{64}{9} - u^2 = \frac{67}{12}
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = \frac{67}{12}-\frac{64}{9} = -\frac{55}{36}
Simplify the expression by subtracting \frac{64}{9} on both sides
u^2 = \frac{55}{36} u = \pm\sqrt{\frac{55}{36}} = \pm \frac{\sqrt{55}}{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{8}{3} - \frac{\sqrt{55}}{6} = 1.431 s = \frac{8}{3} + \frac{\sqrt{55}}{6} = 3.903
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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Limits
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