Skip to main content
Solve for x (complex solution)
Tick mark Image
Graph

Similar Problems from Web Search

Share

-3x^{2}+6x-12=0
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{-6±\sqrt{6^{2}-4\left(-3\right)\left(-12\right)}}{2\left(-3\right)}
This equation is in standard form: ax^{2}+bx+c=0. Substitute -3 for a, 6 for b, and -12 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-6±\sqrt{36-4\left(-3\right)\left(-12\right)}}{2\left(-3\right)}
Square 6.
x=\frac{-6±\sqrt{36+12\left(-12\right)}}{2\left(-3\right)}
Multiply -4 times -3.
x=\frac{-6±\sqrt{36-144}}{2\left(-3\right)}
Multiply 12 times -12.
x=\frac{-6±\sqrt{-108}}{2\left(-3\right)}
Add 36 to -144.
x=\frac{-6±6\sqrt{3}i}{2\left(-3\right)}
Take the square root of -108.
x=\frac{-6±6\sqrt{3}i}{-6}
Multiply 2 times -3.
x=\frac{-6+6\sqrt{3}i}{-6}
Now solve the equation x=\frac{-6±6\sqrt{3}i}{-6} when ± is plus. Add -6 to 6i\sqrt{3}.
x=-\sqrt{3}i+1
Divide -6+6i\sqrt{3} by -6.
x=\frac{-6\sqrt{3}i-6}{-6}
Now solve the equation x=\frac{-6±6\sqrt{3}i}{-6} when ± is minus. Subtract 6i\sqrt{3} from -6.
x=1+\sqrt{3}i
Divide -6-6i\sqrt{3} by -6.
x=-\sqrt{3}i+1 x=1+\sqrt{3}i
The equation is now solved.
-3x^{2}+6x-12=0
Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.
-3x^{2}+6x-12-\left(-12\right)=-\left(-12\right)
Add 12 to both sides of the equation.
-3x^{2}+6x=-\left(-12\right)
Subtracting -12 from itself leaves 0.
-3x^{2}+6x=12
Subtract -12 from 0.
\frac{-3x^{2}+6x}{-3}=\frac{12}{-3}
Divide both sides by -3.
x^{2}+\frac{6}{-3}x=\frac{12}{-3}
Dividing by -3 undoes the multiplication by -3.
x^{2}-2x=\frac{12}{-3}
Divide 6 by -3.
x^{2}-2x=-4
Divide 12 by -3.
x^{2}-2x+1=-4+1
Divide -2, the coefficient of the x term, by 2 to get -1. Then add the square of -1 to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}-2x+1=-3
Add -4 to 1.
\left(x-1\right)^{2}=-3
Factor x^{2}-2x+1. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.
\sqrt{\left(x-1\right)^{2}}=\sqrt{-3}
Take the square root of both sides of the equation.
x-1=\sqrt{3}i x-1=-\sqrt{3}i
Simplify.
x=1+\sqrt{3}i x=-\sqrt{3}i+1
Add 1 to both sides of the equation.
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 = 3
Simplify the expression by subtracting 1 on both sides
u^2 = -3 u = \pm\sqrt{-3} = \pm \sqrt{3}i
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{3}i s = 1 + \sqrt{3}i
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.