Solve for x (complex solution)
x=\frac{3+\sqrt{15}i}{4}\approx 0.75+0.968245837i
x=\frac{-\sqrt{15}i+3}{4}\approx 0.75-0.968245837i
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2x^{2}-3x+3=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 2\times 3}}{2\times 2}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, -3 for b, and 3 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-\left(-3\right)±\sqrt{9-4\times 2\times 3}}{2\times 2}
Square -3.
x=\frac{-\left(-3\right)±\sqrt{9-8\times 3}}{2\times 2}
Multiply -4 times 2.
x=\frac{-\left(-3\right)±\sqrt{9-24}}{2\times 2}
Multiply -8 times 3.
x=\frac{-\left(-3\right)±\sqrt{-15}}{2\times 2}
Add 9 to -24.
x=\frac{-\left(-3\right)±\sqrt{15}i}{2\times 2}
Take the square root of -15.
x=\frac{3±\sqrt{15}i}{2\times 2}
The opposite of -3 is 3.
x=\frac{3±\sqrt{15}i}{4}
Multiply 2 times 2.
x=\frac{3+\sqrt{15}i}{4}
Now solve the equation x=\frac{3±\sqrt{15}i}{4} when ± is plus. Add 3 to i\sqrt{15}.
x=\frac{-\sqrt{15}i+3}{4}
Now solve the equation x=\frac{3±\sqrt{15}i}{4} when ± is minus. Subtract i\sqrt{15} from 3.
x=\frac{3+\sqrt{15}i}{4} x=\frac{-\sqrt{15}i+3}{4}
The equation is now solved.
2x^{2}-3x+3=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.
2x^{2}-3x+3-3=-3
Subtract 3 from both sides of the equation.
2x^{2}-3x=-3
Subtracting 3 from itself leaves 0.
\frac{2x^{2}-3x}{2}=-\frac{3}{2}
Divide both sides by 2.
x^{2}-\frac{3}{2}x=-\frac{3}{2}
Dividing by 2 undoes the multiplication by 2.
x^{2}-\frac{3}{2}x+\left(-\frac{3}{4}\right)^{2}=-\frac{3}{2}+\left(-\frac{3}{4}\right)^{2}
Divide -\frac{3}{2}, the coefficient of the x term, by 2 to get -\frac{3}{4}. Then add the square of -\frac{3}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}-\frac{3}{2}x+\frac{9}{16}=-\frac{3}{2}+\frac{9}{16}
Square -\frac{3}{4} by squaring both the numerator and the denominator of the fraction.
x^{2}-\frac{3}{2}x+\frac{9}{16}=-\frac{15}{16}
Add -\frac{3}{2} to \frac{9}{16} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
\left(x-\frac{3}{4}\right)^{2}=-\frac{15}{16}
Factor x^{2}-\frac{3}{2}x+\frac{9}{16}. 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{3}{4}\right)^{2}}=\sqrt{-\frac{15}{16}}
Take the square root of both sides of the equation.
x-\frac{3}{4}=\frac{\sqrt{15}i}{4} x-\frac{3}{4}=-\frac{\sqrt{15}i}{4}
Simplify.
x=\frac{3+\sqrt{15}i}{4} x=\frac{-\sqrt{15}i+3}{4}
Add \frac{3}{4} to both sides of the equation.
x ^ 2 -\frac{3}{2}x +\frac{3}{2} = 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 2
r + s = \frac{3}{2} rs = \frac{3}{2}
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{3}{4} - u s = \frac{3}{4} + u
Two numbers r and s sum up to \frac{3}{2} exactly when the average of the two numbers is \frac{1}{2}*\frac{3}{2} = \frac{3}{4}. 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{3}{4} - u) (\frac{3}{4} + u) = \frac{3}{2}
To solve for unknown quantity u, substitute these in the product equation rs = \frac{3}{2}
\frac{9}{16} - u^2 = \frac{3}{2}
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = \frac{3}{2}-\frac{9}{16} = \frac{15}{16}
Simplify the expression by subtracting \frac{9}{16} on both sides
u^2 = -\frac{15}{16} u = \pm\sqrt{-\frac{15}{16}} = \pm \frac{\sqrt{15}}{4}i
Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u
r =\frac{3}{4} - \frac{\sqrt{15}}{4}i = 0.750 - 0.968i s = \frac{3}{4} + \frac{\sqrt{15}}{4}i = 0.750 + 0.968i
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
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}