Solve for x
x=\frac{5\sqrt{3}}{3}+3\approx 5.886751346
x=-\frac{5\sqrt{3}}{3}+3\approx 0.113248654
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3x^{2}-18x+2=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{-\left(-18\right)±\sqrt{\left(-18\right)^{2}-4\times 3\times 2}}{2\times 3}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, -18 for b, and 2 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-\left(-18\right)±\sqrt{324-4\times 3\times 2}}{2\times 3}
Square -18.
x=\frac{-\left(-18\right)±\sqrt{324-12\times 2}}{2\times 3}
Multiply -4 times 3.
x=\frac{-\left(-18\right)±\sqrt{324-24}}{2\times 3}
Multiply -12 times 2.
x=\frac{-\left(-18\right)±\sqrt{300}}{2\times 3}
Add 324 to -24.
x=\frac{-\left(-18\right)±10\sqrt{3}}{2\times 3}
Take the square root of 300.
x=\frac{18±10\sqrt{3}}{2\times 3}
The opposite of -18 is 18.
x=\frac{18±10\sqrt{3}}{6}
Multiply 2 times 3.
x=\frac{10\sqrt{3}+18}{6}
Now solve the equation x=\frac{18±10\sqrt{3}}{6} when ± is plus. Add 18 to 10\sqrt{3}.
x=\frac{5\sqrt{3}}{3}+3
Divide 18+10\sqrt{3} by 6.
x=\frac{18-10\sqrt{3}}{6}
Now solve the equation x=\frac{18±10\sqrt{3}}{6} when ± is minus. Subtract 10\sqrt{3} from 18.
x=-\frac{5\sqrt{3}}{3}+3
Divide 18-10\sqrt{3} by 6.
x=\frac{5\sqrt{3}}{3}+3 x=-\frac{5\sqrt{3}}{3}+3
The equation is now solved.
3x^{2}-18x+2=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}-18x+2-2=-2
Subtract 2 from both sides of the equation.
3x^{2}-18x=-2
Subtracting 2 from itself leaves 0.
\frac{3x^{2}-18x}{3}=-\frac{2}{3}
Divide both sides by 3.
x^{2}+\left(-\frac{18}{3}\right)x=-\frac{2}{3}
Dividing by 3 undoes the multiplication by 3.
x^{2}-6x=-\frac{2}{3}
Divide -18 by 3.
x^{2}-6x+\left(-3\right)^{2}=-\frac{2}{3}+\left(-3\right)^{2}
Divide -6, the coefficient of the x term, by 2 to get -3. Then add the square of -3 to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}-6x+9=-\frac{2}{3}+9
Square -3.
x^{2}-6x+9=\frac{25}{3}
Add -\frac{2}{3} to 9.
\left(x-3\right)^{2}=\frac{25}{3}
Factor x^{2}-6x+9. 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-3\right)^{2}}=\sqrt{\frac{25}{3}}
Take the square root of both sides of the equation.
x-3=\frac{5\sqrt{3}}{3} x-3=-\frac{5\sqrt{3}}{3}
Simplify.
x=\frac{5\sqrt{3}}{3}+3 x=-\frac{5\sqrt{3}}{3}+3
Add 3 to both sides of the equation.
x ^ 2 -6x +\frac{2}{3} = 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.This is achieved by dividing both sides of the equation by 3
r + s = 6 rs = \frac{2}{3}
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) = \frac{2}{3}
To solve for unknown quantity u, substitute these in the product equation rs = \frac{2}{3}
9 - u^2 = \frac{2}{3}
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = \frac{2}{3}-9 = -\frac{25}{3}
Simplify the expression by subtracting 9 on both sides
u^2 = \frac{25}{3} u = \pm\sqrt{\frac{25}{3}} = \pm \frac{5}{\sqrt{3}}
Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u
r =3 - \frac{5}{\sqrt{3}} = 0.113 s = 3 + \frac{5}{\sqrt{3}} = 5.887
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}