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12x^{2}+68x+38=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{-68±\sqrt{68^{2}-4\times 12\times 38}}{2\times 12}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 12 for a, 68 for b, and 38 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-68±\sqrt{4624-4\times 12\times 38}}{2\times 12}
Square 68.
x=\frac{-68±\sqrt{4624-48\times 38}}{2\times 12}
Multiply -4 times 12.
x=\frac{-68±\sqrt{4624-1824}}{2\times 12}
Multiply -48 times 38.
x=\frac{-68±\sqrt{2800}}{2\times 12}
Add 4624 to -1824.
x=\frac{-68±20\sqrt{7}}{2\times 12}
Take the square root of 2800.
x=\frac{-68±20\sqrt{7}}{24}
Multiply 2 times 12.
x=\frac{20\sqrt{7}-68}{24}
Now solve the equation x=\frac{-68±20\sqrt{7}}{24} when ± is plus. Add -68 to 20\sqrt{7}.
x=\frac{5\sqrt{7}-17}{6}
Divide -68+20\sqrt{7} by 24.
x=\frac{-20\sqrt{7}-68}{24}
Now solve the equation x=\frac{-68±20\sqrt{7}}{24} when ± is minus. Subtract 20\sqrt{7} from -68.
x=\frac{-5\sqrt{7}-17}{6}
Divide -68-20\sqrt{7} by 24.
x=\frac{5\sqrt{7}-17}{6} x=\frac{-5\sqrt{7}-17}{6}
The equation is now solved.
12x^{2}+68x+38=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}+68x+38-38=-38
Subtract 38 from both sides of the equation.
12x^{2}+68x=-38
Subtracting 38 from itself leaves 0.
\frac{12x^{2}+68x}{12}=-\frac{38}{12}
Divide both sides by 12.
x^{2}+\frac{68}{12}x=-\frac{38}{12}
Dividing by 12 undoes the multiplication by 12.
x^{2}+\frac{17}{3}x=-\frac{38}{12}
Reduce the fraction \frac{68}{12} to lowest terms by extracting and canceling out 4.
x^{2}+\frac{17}{3}x=-\frac{19}{6}
Reduce the fraction \frac{-38}{12} to lowest terms by extracting and canceling out 2.
x^{2}+\frac{17}{3}x+\left(\frac{17}{6}\right)^{2}=-\frac{19}{6}+\left(\frac{17}{6}\right)^{2}
Divide \frac{17}{3}, the coefficient of the x term, by 2 to get \frac{17}{6}. Then add the square of \frac{17}{6} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}+\frac{17}{3}x+\frac{289}{36}=-\frac{19}{6}+\frac{289}{36}
Square \frac{17}{6} by squaring both the numerator and the denominator of the fraction.
x^{2}+\frac{17}{3}x+\frac{289}{36}=\frac{175}{36}
Add -\frac{19}{6} to \frac{289}{36} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
\left(x+\frac{17}{6}\right)^{2}=\frac{175}{36}
Factor x^{2}+\frac{17}{3}x+\frac{289}{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{17}{6}\right)^{2}}=\sqrt{\frac{175}{36}}
Take the square root of both sides of the equation.
x+\frac{17}{6}=\frac{5\sqrt{7}}{6} x+\frac{17}{6}=-\frac{5\sqrt{7}}{6}
Simplify.
x=\frac{5\sqrt{7}-17}{6} x=\frac{-5\sqrt{7}-17}{6}
Subtract \frac{17}{6} from both sides of the equation.
x ^ 2 +\frac{17}{3}x +\frac{19}{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 12
r + s = -\frac{17}{3} rs = \frac{19}{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 = -\frac{17}{6} - u s = -\frac{17}{6} + u
Two numbers r and s sum up to -\frac{17}{3} exactly when the average of the two numbers is \frac{1}{2}*-\frac{17}{3} = -\frac{17}{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{17}{6} - u) (-\frac{17}{6} + u) = \frac{19}{6}
To solve for unknown quantity u, substitute these in the product equation rs = \frac{19}{6}
\frac{289}{36} - u^2 = \frac{19}{6}
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = \frac{19}{6}-\frac{289}{36} = -\frac{175}{36}
Simplify the expression by subtracting \frac{289}{36} on both sides
u^2 = \frac{175}{36} u = \pm\sqrt{\frac{175}{36}} = \pm \frac{\sqrt{175}}{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{17}{6} - \frac{\sqrt{175}}{6} = -5.038 s = -\frac{17}{6} + \frac{\sqrt{175}}{6} = -0.629
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.