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-x^{2}-6x+5=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(-1\right)\times 5}}{2\left(-1\right)}
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(-1\right)\times 5}}{2\left(-1\right)}
Square -6.
x=\frac{-\left(-6\right)±\sqrt{36+4\times 5}}{2\left(-1\right)}
Multiply -4 times -1.
x=\frac{-\left(-6\right)±\sqrt{36+20}}{2\left(-1\right)}
Multiply 4 times 5.
x=\frac{-\left(-6\right)±\sqrt{56}}{2\left(-1\right)}
Add 36 to 20.
x=\frac{-\left(-6\right)±2\sqrt{14}}{2\left(-1\right)}
Take the square root of 56.
x=\frac{6±2\sqrt{14}}{2\left(-1\right)}
The opposite of -6 is 6.
x=\frac{6±2\sqrt{14}}{-2}
Multiply 2 times -1.
x=\frac{2\sqrt{14}+6}{-2}
Now solve the equation x=\frac{6±2\sqrt{14}}{-2} when ± is plus. Add 6 to 2\sqrt{14}.
x=-\left(\sqrt{14}+3\right)
Divide 6+2\sqrt{14} by -2.
x=\frac{6-2\sqrt{14}}{-2}
Now solve the equation x=\frac{6±2\sqrt{14}}{-2} when ± is minus. Subtract 2\sqrt{14} from 6.
x=\sqrt{14}-3
Divide 6-2\sqrt{14} by -2.
-x^{2}-6x+5=-\left(x-\left(-\left(\sqrt{14}+3\right)\right)\right)\left(x-\left(\sqrt{14}-3\right)\right)
Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute -\left(3+\sqrt{14}\right) for x_{1} and -3+\sqrt{14} for x_{2}.
x ^ 2 +6x -5 = 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 = -5
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) = -5
To solve for unknown quantity u, substitute these in the product equation rs = -5
9 - u^2 = -5
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = -5-9 = -14
Simplify the expression by subtracting 9 on both sides
u^2 = 14 u = \pm\sqrt{14} = \pm \sqrt{14}
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{14} = -6.742 s = -3 + \sqrt{14} = 0.742
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.