Solve for x (complex solution)
x=-\frac{5}{2}+\frac{5}{2}i=-2.5+2.5i
x=-\frac{5}{2}-\frac{5}{2}i=-2.5-2.5i
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2x^{2}+10x+25=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{-10±\sqrt{10^{2}-4\times 2\times 25}}{2\times 2}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, 10 for b, and 25 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-10±\sqrt{100-4\times 2\times 25}}{2\times 2}
Square 10.
x=\frac{-10±\sqrt{100-8\times 25}}{2\times 2}
Multiply -4 times 2.
x=\frac{-10±\sqrt{100-200}}{2\times 2}
Multiply -8 times 25.
x=\frac{-10±\sqrt{-100}}{2\times 2}
Add 100 to -200.
x=\frac{-10±10i}{2\times 2}
Take the square root of -100.
x=\frac{-10±10i}{4}
Multiply 2 times 2.
x=\frac{-10+10i}{4}
Now solve the equation x=\frac{-10±10i}{4} when ± is plus. Add -10 to 10i.
x=-\frac{5}{2}+\frac{5}{2}i
Divide -10+10i by 4.
x=\frac{-10-10i}{4}
Now solve the equation x=\frac{-10±10i}{4} when ± is minus. Subtract 10i from -10.
x=-\frac{5}{2}-\frac{5}{2}i
Divide -10-10i by 4.
x=-\frac{5}{2}+\frac{5}{2}i x=-\frac{5}{2}-\frac{5}{2}i
The equation is now solved.
2x^{2}+10x+25=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}+10x+25-25=-25
Subtract 25 from both sides of the equation.
2x^{2}+10x=-25
Subtracting 25 from itself leaves 0.
\frac{2x^{2}+10x}{2}=-\frac{25}{2}
Divide both sides by 2.
x^{2}+\frac{10}{2}x=-\frac{25}{2}
Dividing by 2 undoes the multiplication by 2.
x^{2}+5x=-\frac{25}{2}
Divide 10 by 2.
x^{2}+5x+\left(\frac{5}{2}\right)^{2}=-\frac{25}{2}+\left(\frac{5}{2}\right)^{2}
Divide 5, the coefficient of the x term, by 2 to get \frac{5}{2}. Then add the square of \frac{5}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}+5x+\frac{25}{4}=-\frac{25}{2}+\frac{25}{4}
Square \frac{5}{2} by squaring both the numerator and the denominator of the fraction.
x^{2}+5x+\frac{25}{4}=-\frac{25}{4}
Add -\frac{25}{2} to \frac{25}{4} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
\left(x+\frac{5}{2}\right)^{2}=-\frac{25}{4}
Factor x^{2}+5x+\frac{25}{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{5}{2}\right)^{2}}=\sqrt{-\frac{25}{4}}
Take the square root of both sides of the equation.
x+\frac{5}{2}=\frac{5}{2}i x+\frac{5}{2}=-\frac{5}{2}i
Simplify.
x=-\frac{5}{2}+\frac{5}{2}i x=-\frac{5}{2}-\frac{5}{2}i
Subtract \frac{5}{2} from both sides of the equation.
x ^ 2 +5x +\frac{25}{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 = -5 rs = \frac{25}{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{5}{2} - u s = -\frac{5}{2} + u
Two numbers r and s sum up to -5 exactly when the average of the two numbers is \frac{1}{2}*-5 = -\frac{5}{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{5}{2} - u) (-\frac{5}{2} + u) = \frac{25}{2}
To solve for unknown quantity u, substitute these in the product equation rs = \frac{25}{2}
\frac{25}{4} - u^2 = \frac{25}{2}
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = \frac{25}{2}-\frac{25}{4} = \frac{25}{4}
Simplify the expression by subtracting \frac{25}{4} on both sides
u^2 = -\frac{25}{4} u = \pm\sqrt{-\frac{25}{4}} = \pm \frac{5}{2}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{5}{2} - \frac{5}{2}i = -2.500 - 2.500i s = -\frac{5}{2} + \frac{5}{2}i = -2.500 + 2.500i
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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Matrix
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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
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Limits
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