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3x^{2}+19x+2=0
Quadratic polynomial can be factored using the transformation ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right), where x_{1} and x_{2} are the solutions of the quadratic equation ax^{2}+bx+c=0.
x=\frac{-19±\sqrt{19^{2}-4\times 3\times 2}}{2\times 3}
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{-19±\sqrt{361-4\times 3\times 2}}{2\times 3}
Square 19.
x=\frac{-19±\sqrt{361-12\times 2}}{2\times 3}
Multiply -4 times 3.
x=\frac{-19±\sqrt{361-24}}{2\times 3}
Multiply -12 times 2.
x=\frac{-19±\sqrt{337}}{2\times 3}
Add 361 to -24.
x=\frac{-19±\sqrt{337}}{6}
Multiply 2 times 3.
x=\frac{\sqrt{337}-19}{6}
Now solve the equation x=\frac{-19±\sqrt{337}}{6} when ± is plus. Add -19 to \sqrt{337}.
x=\frac{-\sqrt{337}-19}{6}
Now solve the equation x=\frac{-19±\sqrt{337}}{6} when ± is minus. Subtract \sqrt{337} from -19.
3x^{2}+19x+2=3\left(x-\frac{\sqrt{337}-19}{6}\right)\left(x-\frac{-\sqrt{337}-19}{6}\right)
Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{-19+\sqrt{337}}{6} for x_{1} and \frac{-19-\sqrt{337}}{6} for x_{2}.
x ^ 2 +\frac{19}{3}x +\frac{2}{3} = 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 3
r + s = -\frac{19}{3} rs = \frac{2}{3}
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{19}{6} - u s = -\frac{19}{6} + u
Two numbers r and s sum up to -\frac{19}{3} exactly when the average of the two numbers is \frac{1}{2}*-\frac{19}{3} = -\frac{19}{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{19}{6} - u) (-\frac{19}{6} + u) = \frac{2}{3}
To solve for unknown quantity u, substitute these in the product equation rs = \frac{2}{3}
\frac{361}{36} - u^2 = \frac{2}{3}
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = \frac{2}{3}-\frac{361}{36} = -\frac{337}{36}
Simplify the expression by subtracting \frac{361}{36} on both sides
u^2 = \frac{337}{36} u = \pm\sqrt{\frac{337}{36}} = \pm \frac{\sqrt{337}}{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{19}{6} - \frac{\sqrt{337}}{6} = -6.226 s = -\frac{19}{6} + \frac{\sqrt{337}}{6} = -0.107
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.