Solve for x
x=\frac{\sqrt{33}}{6}+\frac{1}{2}\approx 1.457427108
x=-\frac{\sqrt{33}}{6}+\frac{1}{2}\approx -0.457427108
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3x^{2}-3x-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(-3\right)±\sqrt{\left(-3\right)^{2}-4\times 3\left(-2\right)}}{2\times 3}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, -3 for b, and -2 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-\left(-3\right)±\sqrt{9-4\times 3\left(-2\right)}}{2\times 3}
Square -3.
x=\frac{-\left(-3\right)±\sqrt{9-12\left(-2\right)}}{2\times 3}
Multiply -4 times 3.
x=\frac{-\left(-3\right)±\sqrt{9+24}}{2\times 3}
Multiply -12 times -2.
x=\frac{-\left(-3\right)±\sqrt{33}}{2\times 3}
Add 9 to 24.
x=\frac{3±\sqrt{33}}{2\times 3}
The opposite of -3 is 3.
x=\frac{3±\sqrt{33}}{6}
Multiply 2 times 3.
x=\frac{\sqrt{33}+3}{6}
Now solve the equation x=\frac{3±\sqrt{33}}{6} when ± is plus. Add 3 to \sqrt{33}.
x=\frac{\sqrt{33}}{6}+\frac{1}{2}
Divide 3+\sqrt{33} by 6.
x=\frac{3-\sqrt{33}}{6}
Now solve the equation x=\frac{3±\sqrt{33}}{6} when ± is minus. Subtract \sqrt{33} from 3.
x=-\frac{\sqrt{33}}{6}+\frac{1}{2}
Divide 3-\sqrt{33} by 6.
x=\frac{\sqrt{33}}{6}+\frac{1}{2} x=-\frac{\sqrt{33}}{6}+\frac{1}{2}
The equation is now solved.
3x^{2}-3x-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}-3x-2-\left(-2\right)=-\left(-2\right)
Add 2 to both sides of the equation.
3x^{2}-3x=-\left(-2\right)
Subtracting -2 from itself leaves 0.
3x^{2}-3x=2
Subtract -2 from 0.
\frac{3x^{2}-3x}{3}=\frac{2}{3}
Divide both sides by 3.
x^{2}+\left(-\frac{3}{3}\right)x=\frac{2}{3}
Dividing by 3 undoes the multiplication by 3.
x^{2}-x=\frac{2}{3}
Divide -3 by 3.
x^{2}-x+\left(-\frac{1}{2}\right)^{2}=\frac{2}{3}+\left(-\frac{1}{2}\right)^{2}
Divide -1, the coefficient of the x term, by 2 to get -\frac{1}{2}. Then add the square of -\frac{1}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}-x+\frac{1}{4}=\frac{2}{3}+\frac{1}{4}
Square -\frac{1}{2} by squaring both the numerator and the denominator of the fraction.
x^{2}-x+\frac{1}{4}=\frac{11}{12}
Add \frac{2}{3} to \frac{1}{4} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
\left(x-\frac{1}{2}\right)^{2}=\frac{11}{12}
Factor x^{2}-x+\frac{1}{4}. 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-\frac{1}{2}\right)^{2}}=\sqrt{\frac{11}{12}}
Take the square root of both sides of the equation.
x-\frac{1}{2}=\frac{\sqrt{33}}{6} x-\frac{1}{2}=-\frac{\sqrt{33}}{6}
Simplify.
x=\frac{\sqrt{33}}{6}+\frac{1}{2} x=-\frac{\sqrt{33}}{6}+\frac{1}{2}
Add \frac{1}{2} to both sides of the equation.
x ^ 2 -1x -\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 = 1 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 = \frac{1}{2} - u s = \frac{1}{2} + u
Two numbers r and s sum up to 1 exactly when the average of the two numbers is \frac{1}{2}*1 = \frac{1}{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{1}{2} - u) (\frac{1}{2} + u) = -\frac{2}{3}
To solve for unknown quantity u, substitute these in the product equation rs = -\frac{2}{3}
\frac{1}{4} - u^2 = -\frac{2}{3}
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = -\frac{2}{3}-\frac{1}{4} = -\frac{11}{12}
Simplify the expression by subtracting \frac{1}{4} on both sides
u^2 = \frac{11}{12} u = \pm\sqrt{\frac{11}{12}} = \pm \frac{\sqrt{11}}{\sqrt{12}}
Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u
r =\frac{1}{2} - \frac{\sqrt{11}}{\sqrt{12}} = -0.457 s = \frac{1}{2} + \frac{\sqrt{11}}{\sqrt{12}} = 1.457
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.
Examples
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{ x } ^ { 2 } - 4 x - 5 = 0
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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
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