Solve for x (complex solution)
x=\frac{-7+\sqrt{151}i}{50}\approx -0.14+0.245764115i
x=\frac{-\sqrt{151}i-7}{50}\approx -0.14-0.245764115i
Graph
Share
Copied to clipboard
25x^{2}+7x+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{-7±\sqrt{7^{2}-4\times 25\times 2}}{2\times 25}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 25 for a, 7 for b, and 2 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-7±\sqrt{49-4\times 25\times 2}}{2\times 25}
Square 7.
x=\frac{-7±\sqrt{49-100\times 2}}{2\times 25}
Multiply -4 times 25.
x=\frac{-7±\sqrt{49-200}}{2\times 25}
Multiply -100 times 2.
x=\frac{-7±\sqrt{-151}}{2\times 25}
Add 49 to -200.
x=\frac{-7±\sqrt{151}i}{2\times 25}
Take the square root of -151.
x=\frac{-7±\sqrt{151}i}{50}
Multiply 2 times 25.
x=\frac{-7+\sqrt{151}i}{50}
Now solve the equation x=\frac{-7±\sqrt{151}i}{50} when ± is plus. Add -7 to i\sqrt{151}.
x=\frac{-\sqrt{151}i-7}{50}
Now solve the equation x=\frac{-7±\sqrt{151}i}{50} when ± is minus. Subtract i\sqrt{151} from -7.
x=\frac{-7+\sqrt{151}i}{50} x=\frac{-\sqrt{151}i-7}{50}
The equation is now solved.
25x^{2}+7x+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.
25x^{2}+7x+2-2=-2
Subtract 2 from both sides of the equation.
25x^{2}+7x=-2
Subtracting 2 from itself leaves 0.
\frac{25x^{2}+7x}{25}=-\frac{2}{25}
Divide both sides by 25.
x^{2}+\frac{7}{25}x=-\frac{2}{25}
Dividing by 25 undoes the multiplication by 25.
x^{2}+\frac{7}{25}x+\left(\frac{7}{50}\right)^{2}=-\frac{2}{25}+\left(\frac{7}{50}\right)^{2}
Divide \frac{7}{25}, the coefficient of the x term, by 2 to get \frac{7}{50}. Then add the square of \frac{7}{50} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}+\frac{7}{25}x+\frac{49}{2500}=-\frac{2}{25}+\frac{49}{2500}
Square \frac{7}{50} by squaring both the numerator and the denominator of the fraction.
x^{2}+\frac{7}{25}x+\frac{49}{2500}=-\frac{151}{2500}
Add -\frac{2}{25} to \frac{49}{2500} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
\left(x+\frac{7}{50}\right)^{2}=-\frac{151}{2500}
Factor x^{2}+\frac{7}{25}x+\frac{49}{2500}. 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{7}{50}\right)^{2}}=\sqrt{-\frac{151}{2500}}
Take the square root of both sides of the equation.
x+\frac{7}{50}=\frac{\sqrt{151}i}{50} x+\frac{7}{50}=-\frac{\sqrt{151}i}{50}
Simplify.
x=\frac{-7+\sqrt{151}i}{50} x=\frac{-\sqrt{151}i-7}{50}
Subtract \frac{7}{50} from both sides of the equation.
x ^ 2 +\frac{7}{25}x +\frac{2}{25} = 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 25
r + s = -\frac{7}{25} rs = \frac{2}{25}
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{7}{50} - u s = -\frac{7}{50} + u
Two numbers r and s sum up to -\frac{7}{25} exactly when the average of the two numbers is \frac{1}{2}*-\frac{7}{25} = -\frac{7}{50}. 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{7}{50} - u) (-\frac{7}{50} + u) = \frac{2}{25}
To solve for unknown quantity u, substitute these in the product equation rs = \frac{2}{25}
\frac{49}{2500} - u^2 = \frac{2}{25}
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = \frac{2}{25}-\frac{49}{2500} = \frac{151}{2500}
Simplify the expression by subtracting \frac{49}{2500} on both sides
u^2 = -\frac{151}{2500} u = \pm\sqrt{-\frac{151}{2500}} = \pm \frac{\sqrt{151}}{50}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{7}{50} - \frac{\sqrt{151}}{50}i = -0.140 - 0.246i s = -\frac{7}{50} + \frac{\sqrt{151}}{50}i = -0.140 + 0.246i
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}