Factor
-3\left(x-\left(-\sqrt{5}-1\right)\right)\left(x-\left(\sqrt{5}-1\right)\right)
Evaluate
12-6x-3x^{2}
Graph
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-3x^{2}-6x+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(-6\right)±\sqrt{\left(-6\right)^{2}-4\left(-3\right)\times 12}}{2\left(-3\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(-3\right)\times 12}}{2\left(-3\right)}
Square -6.
x=\frac{-\left(-6\right)±\sqrt{36+12\times 12}}{2\left(-3\right)}
Multiply -4 times -3.
x=\frac{-\left(-6\right)±\sqrt{36+144}}{2\left(-3\right)}
Multiply 12 times 12.
x=\frac{-\left(-6\right)±\sqrt{180}}{2\left(-3\right)}
Add 36 to 144.
x=\frac{-\left(-6\right)±6\sqrt{5}}{2\left(-3\right)}
Take the square root of 180.
x=\frac{6±6\sqrt{5}}{2\left(-3\right)}
The opposite of -6 is 6.
x=\frac{6±6\sqrt{5}}{-6}
Multiply 2 times -3.
x=\frac{6\sqrt{5}+6}{-6}
Now solve the equation x=\frac{6±6\sqrt{5}}{-6} when ± is plus. Add 6 to 6\sqrt{5}.
x=-\left(\sqrt{5}+1\right)
Divide 6+6\sqrt{5} by -6.
x=\frac{6-6\sqrt{5}}{-6}
Now solve the equation x=\frac{6±6\sqrt{5}}{-6} when ± is minus. Subtract 6\sqrt{5} from 6.
x=\sqrt{5}-1
Divide 6-6\sqrt{5} by -6.
-3x^{2}-6x+12=-3\left(x-\left(-\left(\sqrt{5}+1\right)\right)\right)\left(x-\left(\sqrt{5}-1\right)\right)
Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute -\left(1+\sqrt{5}\right) for x_{1} and -1+\sqrt{5} for x_{2}.
x ^ 2 +2x -4 = 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 = -2 rs = -4
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 = -1 - u s = -1 + u
Two numbers r and s sum up to -2 exactly when the average of the two numbers is \frac{1}{2}*-2 = -1. 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>
(-1 - u) (-1 + u) = -4
To solve for unknown quantity u, substitute these in the product equation rs = -4
1 - u^2 = -4
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = -4-1 = -5
Simplify the expression by subtracting 1 on both sides
u^2 = 5 u = \pm\sqrt{5} = \pm \sqrt{5}
Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u
r =-1 - \sqrt{5} = -3.236 s = -1 + \sqrt{5} = 1.236
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.
Examples
Quadratic equation
{ x } ^ { 2 } - 4 x - 5 = 0
Trigonometry
4 \sin \theta \cos \theta = 2 \sin \theta
Linear equation
y = 3x + 4
Arithmetic
699 * 533
Matrix
\left[ \begin{array} { l l } { 2 } & { 3 } \\ { 5 } & { 4 } \end{array} \right] \left[ \begin{array} { l l l } { 2 } & { 0 } & { 3 } \\ { -1 } & { 1 } & { 5 } \end{array} \right]
Simultaneous equation
\left. \begin{cases} { 8x+2y = 46 } \\ { 7x+3y = 47 } \end{cases} \right.
Differentiation
\frac { d } { d x } \frac { ( 3 x ^ { 2 } - 2 ) } { ( x - 5 ) }
Integration
\int _ { 0 } ^ { 1 } x e ^ { - x ^ { 2 } } d x
Limits
\lim _{x \rightarrow-3} \frac{x^{2}-9}{x^{2}+2 x-3}