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-y^{2}+10y-504=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.
y=\frac{-10±\sqrt{10^{2}-4\left(-1\right)\left(-504\right)}}{2\left(-1\right)}
This equation is in standard form: ax^{2}+bx+c=0. Substitute -1 for a, 10 for b, and -504 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
y=\frac{-10±\sqrt{100-4\left(-1\right)\left(-504\right)}}{2\left(-1\right)}
Square 10.
y=\frac{-10±\sqrt{100+4\left(-504\right)}}{2\left(-1\right)}
Multiply -4 times -1.
y=\frac{-10±\sqrt{100-2016}}{2\left(-1\right)}
Multiply 4 times -504.
y=\frac{-10±\sqrt{-1916}}{2\left(-1\right)}
Add 100 to -2016.
y=\frac{-10±2\sqrt{479}i}{2\left(-1\right)}
Take the square root of -1916.
y=\frac{-10±2\sqrt{479}i}{-2}
Multiply 2 times -1.
y=\frac{-10+2\sqrt{479}i}{-2}
Now solve the equation y=\frac{-10±2\sqrt{479}i}{-2} when ± is plus. Add -10 to 2i\sqrt{479}.
y=-\sqrt{479}i+5
Divide -10+2i\sqrt{479} by -2.
y=\frac{-2\sqrt{479}i-10}{-2}
Now solve the equation y=\frac{-10±2\sqrt{479}i}{-2} when ± is minus. Subtract 2i\sqrt{479} from -10.
y=5+\sqrt{479}i
Divide -10-2i\sqrt{479} by -2.
y=-\sqrt{479}i+5 y=5+\sqrt{479}i
The equation is now solved.
-y^{2}+10y-504=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.
-y^{2}+10y-504-\left(-504\right)=-\left(-504\right)
Add 504 to both sides of the equation.
-y^{2}+10y=-\left(-504\right)
Subtracting -504 from itself leaves 0.
-y^{2}+10y=504
Subtract -504 from 0.
\frac{-y^{2}+10y}{-1}=\frac{504}{-1}
Divide both sides by -1.
y^{2}+\frac{10}{-1}y=\frac{504}{-1}
Dividing by -1 undoes the multiplication by -1.
y^{2}-10y=\frac{504}{-1}
Divide 10 by -1.
y^{2}-10y=-504
Divide 504 by -1.
y^{2}-10y+\left(-5\right)^{2}=-504+\left(-5\right)^{2}
Divide -10, the coefficient of the x term, by 2 to get -5. Then add the square of -5 to both sides of the equation. This step makes the left hand side of the equation a perfect square.
y^{2}-10y+25=-504+25
Square -5.
y^{2}-10y+25=-479
Add -504 to 25.
\left(y-5\right)^{2}=-479
Factor y^{2}-10y+25. 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(y-5\right)^{2}}=\sqrt{-479}
Take the square root of both sides of the equation.
y-5=\sqrt{479}i y-5=-\sqrt{479}i
Simplify.
y=5+\sqrt{479}i y=-\sqrt{479}i+5
Add 5 to both sides of the equation.
x ^ 2 -10x +504 = 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 = 10 rs = 504
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 = 5 - u s = 5 + u
Two numbers r and s sum up to 10 exactly when the average of the two numbers is \frac{1}{2}*10 = 5. 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>
(5 - u) (5 + u) = 504
To solve for unknown quantity u, substitute these in the product equation rs = 504
25 - u^2 = 504
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = 504-25 = 479
Simplify the expression by subtracting 25 on both sides
u^2 = -479 u = \pm\sqrt{-479} = \pm \sqrt{479}i
Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u
r =5 - \sqrt{479}i s = 5 + \sqrt{479}i
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.