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x^{2}-6x-14=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{-\left(-6\right)±\sqrt{\left(-6\right)^{2}-4\left(-14\right)}}{2}
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(-6\right)±\sqrt{36-4\left(-14\right)}}{2}
Square -6.
x=\frac{-\left(-6\right)±\sqrt{36+56}}{2}
Multiply -4 times -14.
x=\frac{-\left(-6\right)±\sqrt{92}}{2}
Add 36 to 56.
x=\frac{-\left(-6\right)±2\sqrt{23}}{2}
Take the square root of 92.
x=\frac{6±2\sqrt{23}}{2}
The opposite of -6 is 6.
x=\frac{2\sqrt{23}+6}{2}
Now solve the equation x=\frac{6±2\sqrt{23}}{2} when ± is plus. Add 6 to 2\sqrt{23}.
x=\sqrt{23}+3
Divide 6+2\sqrt{23} by 2.
x=\frac{6-2\sqrt{23}}{2}
Now solve the equation x=\frac{6±2\sqrt{23}}{2} when ± is minus. Subtract 2\sqrt{23} from 6.
x=3-\sqrt{23}
Divide 6-2\sqrt{23} by 2.
x^{2}-6x-14=\left(x-\left(\sqrt{23}+3\right)\right)\left(x-\left(3-\sqrt{23}\right)\right)
Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 3+\sqrt{23} for x_{1} and 3-\sqrt{23} for x_{2}.
x ^ 2 -6x -14 = 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.
r + s = 6 rs = -14
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 = 3 - u s = 3 + u
Two numbers r and s sum up to 6 exactly when the average of the two numbers is \frac{1}{2}*6 = 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>
(3 - u) (3 + u) = -14
To solve for unknown quantity u, substitute these in the product equation rs = -14
9 - u^2 = -14
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = -14-9 = -23
Simplify the expression by subtracting 9 on both sides
u^2 = 23 u = \pm\sqrt{23} = \pm \sqrt{23}
Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u
r =3 - \sqrt{23} = -1.796 s = 3 + \sqrt{23} = 7.796
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.