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x^{2}-5x+49=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(-5\right)±\sqrt{\left(-5\right)^{2}-4\times 49}}{2}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -5 for b, and 49 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-\left(-5\right)±\sqrt{25-4\times 49}}{2}
Square -5.
x=\frac{-\left(-5\right)±\sqrt{25-196}}{2}
Multiply -4 times 49.
x=\frac{-\left(-5\right)±\sqrt{-171}}{2}
Add 25 to -196.
x=\frac{-\left(-5\right)±3\sqrt{19}i}{2}
Take the square root of -171.
x=\frac{5±3\sqrt{19}i}{2}
The opposite of -5 is 5.
x=\frac{5+3\sqrt{19}i}{2}
Now solve the equation x=\frac{5±3\sqrt{19}i}{2} when ± is plus. Add 5 to 3i\sqrt{19}.
x=\frac{-3\sqrt{19}i+5}{2}
Now solve the equation x=\frac{5±3\sqrt{19}i}{2} when ± is minus. Subtract 3i\sqrt{19} from 5.
x=\frac{5+3\sqrt{19}i}{2} x=\frac{-3\sqrt{19}i+5}{2}
The equation is now solved.
x^{2}-5x+49=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}-5x+49-49=-49
Subtract 49 from both sides of the equation.
x^{2}-5x=-49
Subtracting 49 from itself leaves 0.
x^{2}-5x+\left(-\frac{5}{2}\right)^{2}=-49+\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}=-49+\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{171}{4}
Add -49 to \frac{25}{4}.
\left(x-\frac{5}{2}\right)^{2}=-\frac{171}{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{171}{4}}
Take the square root of both sides of the equation.
x-\frac{5}{2}=\frac{3\sqrt{19}i}{2} x-\frac{5}{2}=-\frac{3\sqrt{19}i}{2}
Simplify.
x=\frac{5+3\sqrt{19}i}{2} x=\frac{-3\sqrt{19}i+5}{2}
Add \frac{5}{2} to both sides of the equation.
x ^ 2 -5x +49 = 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 = 5 rs = 49
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) = 49
To solve for unknown quantity u, substitute these in the product equation rs = 49
\frac{25}{4} - u^2 = 49
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = 49-\frac{25}{4} = \frac{171}{4}
Simplify the expression by subtracting \frac{25}{4} on both sides
u^2 = -\frac{171}{4} u = \pm\sqrt{-\frac{171}{4}} = \pm \frac{\sqrt{171}}{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{\sqrt{171}}{2}i = 2.500 - 6.538i s = \frac{5}{2} + \frac{\sqrt{171}}{2}i = 2.500 + 6.538i
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.