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4z^{2}-4z+13=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.
z=\frac{-\left(-4\right)±\sqrt{\left(-4\right)^{2}-4\times 4\times 13}}{2\times 4}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 4 for a, -4 for b, and 13 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
z=\frac{-\left(-4\right)±\sqrt{16-4\times 4\times 13}}{2\times 4}
Square -4.
z=\frac{-\left(-4\right)±\sqrt{16-16\times 13}}{2\times 4}
Multiply -4 times 4.
z=\frac{-\left(-4\right)±\sqrt{16-208}}{2\times 4}
Multiply -16 times 13.
z=\frac{-\left(-4\right)±\sqrt{-192}}{2\times 4}
Add 16 to -208.
z=\frac{-\left(-4\right)±8\sqrt{3}i}{2\times 4}
Take the square root of -192.
z=\frac{4±8\sqrt{3}i}{2\times 4}
The opposite of -4 is 4.
z=\frac{4±8\sqrt{3}i}{8}
Multiply 2 times 4.
z=\frac{4+8\sqrt{3}i}{8}
Now solve the equation z=\frac{4±8\sqrt{3}i}{8} when ± is plus. Add 4 to 8i\sqrt{3}.
z=\frac{1}{2}+\sqrt{3}i
Divide 4+8i\sqrt{3} by 8.
z=\frac{-8\sqrt{3}i+4}{8}
Now solve the equation z=\frac{4±8\sqrt{3}i}{8} when ± is minus. Subtract 8i\sqrt{3} from 4.
z=-\sqrt{3}i+\frac{1}{2}
Divide 4-8i\sqrt{3} by 8.
z=\frac{1}{2}+\sqrt{3}i z=-\sqrt{3}i+\frac{1}{2}
The equation is now solved.
4z^{2}-4z+13=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.
4z^{2}-4z+13-13=-13
Subtract 13 from both sides of the equation.
4z^{2}-4z=-13
Subtracting 13 from itself leaves 0.
\frac{4z^{2}-4z}{4}=-\frac{13}{4}
Divide both sides by 4.
z^{2}+\left(-\frac{4}{4}\right)z=-\frac{13}{4}
Dividing by 4 undoes the multiplication by 4.
z^{2}-z=-\frac{13}{4}
Divide -4 by 4.
z^{2}-z+\left(-\frac{1}{2}\right)^{2}=-\frac{13}{4}+\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.
z^{2}-z+\frac{1}{4}=\frac{-13+1}{4}
Square -\frac{1}{2} by squaring both the numerator and the denominator of the fraction.
z^{2}-z+\frac{1}{4}=-3
Add -\frac{13}{4} to \frac{1}{4} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
\left(z-\frac{1}{2}\right)^{2}=-3
Factor z^{2}-z+\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(z-\frac{1}{2}\right)^{2}}=\sqrt{-3}
Take the square root of both sides of the equation.
z-\frac{1}{2}=\sqrt{3}i z-\frac{1}{2}=-\sqrt{3}i
Simplify.
z=\frac{1}{2}+\sqrt{3}i z=-\sqrt{3}i+\frac{1}{2}
Add \frac{1}{2} to both sides of the equation.
x ^ 2 -1x +\frac{13}{4} = 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 4
r + s = 1 rs = \frac{13}{4}
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) = \frac{13}{4}
To solve for unknown quantity u, substitute these in the product equation rs = \frac{13}{4}
\frac{1}{4} - u^2 = \frac{13}{4}
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = \frac{13}{4}-\frac{1}{4} = 3
Simplify the expression by subtracting \frac{1}{4} on both sides
u^2 = -3 u = \pm\sqrt{-3} = \pm \sqrt{3}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} - \sqrt{3}i s = \frac{1}{2} + \sqrt{3}i
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.