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x^{2}-150x+594=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(-150\right)±\sqrt{\left(-150\right)^{2}-4\times 594}}{2}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -150 for b, and 594 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-\left(-150\right)±\sqrt{22500-4\times 594}}{2}
Square -150.
x=\frac{-\left(-150\right)±\sqrt{22500-2376}}{2}
Multiply -4 times 594.
x=\frac{-\left(-150\right)±\sqrt{20124}}{2}
Add 22500 to -2376.
x=\frac{-\left(-150\right)±6\sqrt{559}}{2}
Take the square root of 20124.
x=\frac{150±6\sqrt{559}}{2}
The opposite of -150 is 150.
x=\frac{6\sqrt{559}+150}{2}
Now solve the equation x=\frac{150±6\sqrt{559}}{2} when ± is plus. Add 150 to 6\sqrt{559}.
x=3\sqrt{559}+75
Divide 150+6\sqrt{559} by 2.
x=\frac{150-6\sqrt{559}}{2}
Now solve the equation x=\frac{150±6\sqrt{559}}{2} when ± is minus. Subtract 6\sqrt{559} from 150.
x=75-3\sqrt{559}
Divide 150-6\sqrt{559} by 2.
x=3\sqrt{559}+75 x=75-3\sqrt{559}
The equation is now solved.
x^{2}-150x+594=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}-150x+594-594=-594
Subtract 594 from both sides of the equation.
x^{2}-150x=-594
Subtracting 594 from itself leaves 0.
x^{2}-150x+\left(-75\right)^{2}=-594+\left(-75\right)^{2}
Divide -150, the coefficient of the x term, by 2 to get -75. Then add the square of -75 to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}-150x+5625=-594+5625
Square -75.
x^{2}-150x+5625=5031
Add -594 to 5625.
\left(x-75\right)^{2}=5031
Factor x^{2}-150x+5625. 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-75\right)^{2}}=\sqrt{5031}
Take the square root of both sides of the equation.
x-75=3\sqrt{559} x-75=-3\sqrt{559}
Simplify.
x=3\sqrt{559}+75 x=75-3\sqrt{559}
Add 75 to both sides of the equation.
x ^ 2 -150x +594 = 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 = 150 rs = 594
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 = 75 - u s = 75 + u
Two numbers r and s sum up to 150 exactly when the average of the two numbers is \frac{1}{2}*150 = 75. 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>
(75 - u) (75 + u) = 594
To solve for unknown quantity u, substitute these in the product equation rs = 594
5625 - u^2 = 594
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = 594-5625 = -5031
Simplify the expression by subtracting 5625 on both sides
u^2 = 5031 u = \pm\sqrt{5031} = \pm \sqrt{5031}
Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u
r =75 - \sqrt{5031} = 4.070 s = 75 + \sqrt{5031} = 145.930
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.