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-x^{2}-5x+12=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(-5\right)±\sqrt{\left(-5\right)^{2}-4\left(-1\right)\times 12}}{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(-5\right)±\sqrt{25-4\left(-1\right)\times 12}}{2\left(-1\right)}
Square -5.
x=\frac{-\left(-5\right)±\sqrt{25+4\times 12}}{2\left(-1\right)}
Multiply -4 times -1.
x=\frac{-\left(-5\right)±\sqrt{25+48}}{2\left(-1\right)}
Multiply 4 times 12.
x=\frac{-\left(-5\right)±\sqrt{73}}{2\left(-1\right)}
Add 25 to 48.
x=\frac{5±\sqrt{73}}{2\left(-1\right)}
The opposite of -5 is 5.
x=\frac{5±\sqrt{73}}{-2}
Multiply 2 times -1.
x=\frac{\sqrt{73}+5}{-2}
Now solve the equation x=\frac{5±\sqrt{73}}{-2} when ± is plus. Add 5 to \sqrt{73}.
x=\frac{-\sqrt{73}-5}{2}
Divide 5+\sqrt{73} by -2.
x=\frac{5-\sqrt{73}}{-2}
Now solve the equation x=\frac{5±\sqrt{73}}{-2} when ± is minus. Subtract \sqrt{73} from 5.
x=\frac{\sqrt{73}-5}{2}
Divide 5-\sqrt{73} by -2.
-x^{2}-5x+12=-\left(x-\frac{-\sqrt{73}-5}{2}\right)\left(x-\frac{\sqrt{73}-5}{2}\right)
Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{-5-\sqrt{73}}{2} for x_{1} and \frac{-5+\sqrt{73}}{2} for x_{2}.
x ^ 2 +5x -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.
r + s = -5 rs = -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{5}{2} - u s = -\frac{5}{2} + u
Two numbers r and s sum up to -5 exactly when the average of the two numbers is \frac{1}{2}*-5 = -\frac{5}{2}. 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{5}{2} - u) (-\frac{5}{2} + u) = -12
To solve for unknown quantity u, substitute these in the product equation rs = -12
\frac{25}{4} - u^2 = -12
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = -12-\frac{25}{4} = -\frac{73}{4}
Simplify the expression by subtracting \frac{25}{4} on both sides
u^2 = \frac{73}{4} u = \pm\sqrt{\frac{73}{4}} = \pm \frac{\sqrt{73}}{2}
Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u
r =-\frac{5}{2} - \frac{\sqrt{73}}{2} = -6.772 s = -\frac{5}{2} + \frac{\sqrt{73}}{2} = 1.772
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.