Solve for x (complex solution)
x=\frac{1+\sqrt{14}i}{5}\approx 0.2+0.748331477i
x=\frac{-\sqrt{14}i+1}{5}\approx 0.2-0.748331477i
Graph
Share
Copied to clipboard
5x^{2}-2x+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(-2\right)±\sqrt{\left(-2\right)^{2}-4\times 5\times 3}}{2\times 5}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 5 for a, -2 for b, and 3 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-\left(-2\right)±\sqrt{4-4\times 5\times 3}}{2\times 5}
Square -2.
x=\frac{-\left(-2\right)±\sqrt{4-20\times 3}}{2\times 5}
Multiply -4 times 5.
x=\frac{-\left(-2\right)±\sqrt{4-60}}{2\times 5}
Multiply -20 times 3.
x=\frac{-\left(-2\right)±\sqrt{-56}}{2\times 5}
Add 4 to -60.
x=\frac{-\left(-2\right)±2\sqrt{14}i}{2\times 5}
Take the square root of -56.
x=\frac{2±2\sqrt{14}i}{2\times 5}
The opposite of -2 is 2.
x=\frac{2±2\sqrt{14}i}{10}
Multiply 2 times 5.
x=\frac{2+2\sqrt{14}i}{10}
Now solve the equation x=\frac{2±2\sqrt{14}i}{10} when ± is plus. Add 2 to 2i\sqrt{14}.
x=\frac{1+\sqrt{14}i}{5}
Divide 2+2i\sqrt{14} by 10.
x=\frac{-2\sqrt{14}i+2}{10}
Now solve the equation x=\frac{2±2\sqrt{14}i}{10} when ± is minus. Subtract 2i\sqrt{14} from 2.
x=\frac{-\sqrt{14}i+1}{5}
Divide 2-2i\sqrt{14} by 10.
x=\frac{1+\sqrt{14}i}{5} x=\frac{-\sqrt{14}i+1}{5}
The equation is now solved.
5x^{2}-2x+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.
5x^{2}-2x+3-3=-3
Subtract 3 from both sides of the equation.
5x^{2}-2x=-3
Subtracting 3 from itself leaves 0.
\frac{5x^{2}-2x}{5}=-\frac{3}{5}
Divide both sides by 5.
x^{2}-\frac{2}{5}x=-\frac{3}{5}
Dividing by 5 undoes the multiplication by 5.
x^{2}-\frac{2}{5}x+\left(-\frac{1}{5}\right)^{2}=-\frac{3}{5}+\left(-\frac{1}{5}\right)^{2}
Divide -\frac{2}{5}, the coefficient of the x term, by 2 to get -\frac{1}{5}. Then add the square of -\frac{1}{5} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}-\frac{2}{5}x+\frac{1}{25}=-\frac{3}{5}+\frac{1}{25}
Square -\frac{1}{5} by squaring both the numerator and the denominator of the fraction.
x^{2}-\frac{2}{5}x+\frac{1}{25}=-\frac{14}{25}
Add -\frac{3}{5} to \frac{1}{25} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
\left(x-\frac{1}{5}\right)^{2}=-\frac{14}{25}
Factor x^{2}-\frac{2}{5}x+\frac{1}{25}. 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}{5}\right)^{2}}=\sqrt{-\frac{14}{25}}
Take the square root of both sides of the equation.
x-\frac{1}{5}=\frac{\sqrt{14}i}{5} x-\frac{1}{5}=-\frac{\sqrt{14}i}{5}
Simplify.
x=\frac{1+\sqrt{14}i}{5} x=\frac{-\sqrt{14}i+1}{5}
Add \frac{1}{5} to both sides of the equation.
x ^ 2 -\frac{2}{5}x +\frac{3}{5} = 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 5
r + s = \frac{2}{5} rs = \frac{3}{5}
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}{5} - u s = \frac{1}{5} + u
Two numbers r and s sum up to \frac{2}{5} exactly when the average of the two numbers is \frac{1}{2}*\frac{2}{5} = \frac{1}{5}. 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}{5} - u) (\frac{1}{5} + u) = \frac{3}{5}
To solve for unknown quantity u, substitute these in the product equation rs = \frac{3}{5}
\frac{1}{25} - u^2 = \frac{3}{5}
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = \frac{3}{5}-\frac{1}{25} = \frac{14}{25}
Simplify the expression by subtracting \frac{1}{25} on both sides
u^2 = -\frac{14}{25} u = \pm\sqrt{-\frac{14}{25}} = \pm \frac{\sqrt{14}}{5}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{1}{5} - \frac{\sqrt{14}}{5}i = 0.200 - 0.748i s = \frac{1}{5} + \frac{\sqrt{14}}{5}i = 0.200 + 0.748i
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}