Solve for x (complex solution)
x=\frac{3+3\sqrt{23}i}{2}\approx 1.5+7.193747285i
x=\frac{-3\sqrt{23}i+3}{2}\approx 1.5-7.193747285i
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x^{2}-3x+54=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(-3\right)±\sqrt{\left(-3\right)^{2}-4\times 54}}{2}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -3 for b, and 54 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-\left(-3\right)±\sqrt{9-4\times 54}}{2}
Square -3.
x=\frac{-\left(-3\right)±\sqrt{9-216}}{2}
Multiply -4 times 54.
x=\frac{-\left(-3\right)±\sqrt{-207}}{2}
Add 9 to -216.
x=\frac{-\left(-3\right)±3\sqrt{23}i}{2}
Take the square root of -207.
x=\frac{3±3\sqrt{23}i}{2}
The opposite of -3 is 3.
x=\frac{3+3\sqrt{23}i}{2}
Now solve the equation x=\frac{3±3\sqrt{23}i}{2} when ± is plus. Add 3 to 3i\sqrt{23}.
x=\frac{-3\sqrt{23}i+3}{2}
Now solve the equation x=\frac{3±3\sqrt{23}i}{2} when ± is minus. Subtract 3i\sqrt{23} from 3.
x=\frac{3+3\sqrt{23}i}{2} x=\frac{-3\sqrt{23}i+3}{2}
The equation is now solved.
x^{2}-3x+54=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}-3x+54-54=-54
Subtract 54 from both sides of the equation.
x^{2}-3x=-54
Subtracting 54 from itself leaves 0.
x^{2}-3x+\left(-\frac{3}{2}\right)^{2}=-54+\left(-\frac{3}{2}\right)^{2}
Divide -3, the coefficient of the x term, by 2 to get -\frac{3}{2}. Then add the square of -\frac{3}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}-3x+\frac{9}{4}=-54+\frac{9}{4}
Square -\frac{3}{2} by squaring both the numerator and the denominator of the fraction.
x^{2}-3x+\frac{9}{4}=-\frac{207}{4}
Add -54 to \frac{9}{4}.
\left(x-\frac{3}{2}\right)^{2}=-\frac{207}{4}
Factor x^{2}-3x+\frac{9}{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{3}{2}\right)^{2}}=\sqrt{-\frac{207}{4}}
Take the square root of both sides of the equation.
x-\frac{3}{2}=\frac{3\sqrt{23}i}{2} x-\frac{3}{2}=-\frac{3\sqrt{23}i}{2}
Simplify.
x=\frac{3+3\sqrt{23}i}{2} x=\frac{-3\sqrt{23}i+3}{2}
Add \frac{3}{2} to both sides of the equation.
x ^ 2 -3x +54 = 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 = 3 rs = 54
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{3}{2} - u s = \frac{3}{2} + u
Two numbers r and s sum up to 3 exactly when the average of the two numbers is \frac{1}{2}*3 = \frac{3}{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{3}{2} - u) (\frac{3}{2} + u) = 54
To solve for unknown quantity u, substitute these in the product equation rs = 54
\frac{9}{4} - u^2 = 54
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = 54-\frac{9}{4} = \frac{207}{4}
Simplify the expression by subtracting \frac{9}{4} on both sides
u^2 = -\frac{207}{4} u = \pm\sqrt{-\frac{207}{4}} = \pm \frac{\sqrt{207}}{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{3}{2} - \frac{\sqrt{207}}{2}i = 1.500 - 7.194i s = \frac{3}{2} + \frac{\sqrt{207}}{2}i = 1.500 + 7.194i
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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