Solve for x (complex solution)
x=\frac{-\sqrt{19}i+1}{2}\approx 0.5-2.179449472i
x=\frac{1+\sqrt{19}i}{2}\approx 0.5+2.179449472i
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-x^{2}+x-5=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{-1±\sqrt{1^{2}-4\left(-1\right)\left(-5\right)}}{2\left(-1\right)}
This equation is in standard form: ax^{2}+bx+c=0. Substitute -1 for a, 1 for b, and -5 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-1±\sqrt{1-4\left(-1\right)\left(-5\right)}}{2\left(-1\right)}
Square 1.
x=\frac{-1±\sqrt{1+4\left(-5\right)}}{2\left(-1\right)}
Multiply -4 times -1.
x=\frac{-1±\sqrt{1-20}}{2\left(-1\right)}
Multiply 4 times -5.
x=\frac{-1±\sqrt{-19}}{2\left(-1\right)}
Add 1 to -20.
x=\frac{-1±\sqrt{19}i}{2\left(-1\right)}
Take the square root of -19.
x=\frac{-1±\sqrt{19}i}{-2}
Multiply 2 times -1.
x=\frac{-1+\sqrt{19}i}{-2}
Now solve the equation x=\frac{-1±\sqrt{19}i}{-2} when ± is plus. Add -1 to i\sqrt{19}.
x=\frac{-\sqrt{19}i+1}{2}
Divide -1+i\sqrt{19} by -2.
x=\frac{-\sqrt{19}i-1}{-2}
Now solve the equation x=\frac{-1±\sqrt{19}i}{-2} when ± is minus. Subtract i\sqrt{19} from -1.
x=\frac{1+\sqrt{19}i}{2}
Divide -1-i\sqrt{19} by -2.
x=\frac{-\sqrt{19}i+1}{2} x=\frac{1+\sqrt{19}i}{2}
The equation is now solved.
-x^{2}+x-5=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.
-x^{2}+x-5-\left(-5\right)=-\left(-5\right)
Add 5 to both sides of the equation.
-x^{2}+x=-\left(-5\right)
Subtracting -5 from itself leaves 0.
-x^{2}+x=5
Subtract -5 from 0.
\frac{-x^{2}+x}{-1}=\frac{5}{-1}
Divide both sides by -1.
x^{2}+\frac{1}{-1}x=\frac{5}{-1}
Dividing by -1 undoes the multiplication by -1.
x^{2}-x=\frac{5}{-1}
Divide 1 by -1.
x^{2}-x=-5
Divide 5 by -1.
x^{2}-x+\left(-\frac{1}{2}\right)^{2}=-5+\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}=-5+\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{19}{4}
Add -5 to \frac{1}{4}.
\left(x-\frac{1}{2}\right)^{2}=-\frac{19}{4}
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{19}{4}}
Take the square root of both sides of the equation.
x-\frac{1}{2}=\frac{\sqrt{19}i}{2} x-\frac{1}{2}=-\frac{\sqrt{19}i}{2}
Simplify.
x=\frac{1+\sqrt{19}i}{2} x=\frac{-\sqrt{19}i+1}{2}
Add \frac{1}{2} to both sides of the equation.
x ^ 2 -1x +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.
r + s = 1 rs = 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}{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) = 5
To solve for unknown quantity u, substitute these in the product equation rs = 5
\frac{1}{4} - u^2 = 5
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = 5-\frac{1}{4} = \frac{19}{4}
Simplify the expression by subtracting \frac{1}{4} on both sides
u^2 = -\frac{19}{4} u = \pm\sqrt{-\frac{19}{4}} = \pm \frac{\sqrt{19}}{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{1}{2} - \frac{\sqrt{19}}{2}i = 0.500 - 2.179i s = \frac{1}{2} + \frac{\sqrt{19}}{2}i = 0.500 + 2.179i
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.
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