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-2x^{2}+200x-4350=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{-200±\sqrt{200^{2}-4\left(-2\right)\left(-4350\right)}}{2\left(-2\right)}
This equation is in standard form: ax^{2}+bx+c=0. Substitute -2 for a, 200 for b, and -4350 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-200±\sqrt{40000-4\left(-2\right)\left(-4350\right)}}{2\left(-2\right)}
Square 200.
x=\frac{-200±\sqrt{40000+8\left(-4350\right)}}{2\left(-2\right)}
Multiply -4 times -2.
x=\frac{-200±\sqrt{40000-34800}}{2\left(-2\right)}
Multiply 8 times -4350.
x=\frac{-200±\sqrt{5200}}{2\left(-2\right)}
Add 40000 to -34800.
x=\frac{-200±20\sqrt{13}}{2\left(-2\right)}
Take the square root of 5200.
x=\frac{-200±20\sqrt{13}}{-4}
Multiply 2 times -2.
x=\frac{20\sqrt{13}-200}{-4}
Now solve the equation x=\frac{-200±20\sqrt{13}}{-4} when ± is plus. Add -200 to 20\sqrt{13}.
x=50-5\sqrt{13}
Divide -200+20\sqrt{13} by -4.
x=\frac{-20\sqrt{13}-200}{-4}
Now solve the equation x=\frac{-200±20\sqrt{13}}{-4} when ± is minus. Subtract 20\sqrt{13} from -200.
x=5\sqrt{13}+50
Divide -200-20\sqrt{13} by -4.
x=50-5\sqrt{13} x=5\sqrt{13}+50
The equation is now solved.
-2x^{2}+200x-4350=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.
-2x^{2}+200x-4350-\left(-4350\right)=-\left(-4350\right)
Add 4350 to both sides of the equation.
-2x^{2}+200x=-\left(-4350\right)
Subtracting -4350 from itself leaves 0.
-2x^{2}+200x=4350
Subtract -4350 from 0.
\frac{-2x^{2}+200x}{-2}=\frac{4350}{-2}
Divide both sides by -2.
x^{2}+\frac{200}{-2}x=\frac{4350}{-2}
Dividing by -2 undoes the multiplication by -2.
x^{2}-100x=\frac{4350}{-2}
Divide 200 by -2.
x^{2}-100x=-2175
Divide 4350 by -2.
x^{2}-100x+\left(-50\right)^{2}=-2175+\left(-50\right)^{2}
Divide -100, the coefficient of the x term, by 2 to get -50. Then add the square of -50 to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}-100x+2500=-2175+2500
Square -50.
x^{2}-100x+2500=325
Add -2175 to 2500.
\left(x-50\right)^{2}=325
Factor x^{2}-100x+2500. 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-50\right)^{2}}=\sqrt{325}
Take the square root of both sides of the equation.
x-50=5\sqrt{13} x-50=-5\sqrt{13}
Simplify.
x=5\sqrt{13}+50 x=50-5\sqrt{13}
Add 50 to both sides of the equation.
x ^ 2 -100x +2175 = 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 = 100 rs = 2175
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 = 50 - u s = 50 + u
Two numbers r and s sum up to 100 exactly when the average of the two numbers is \frac{1}{2}*100 = 50. 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>
(50 - u) (50 + u) = 2175
To solve for unknown quantity u, substitute these in the product equation rs = 2175
2500 - u^2 = 2175
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = 2175-2500 = -325
Simplify the expression by subtracting 2500 on both sides
u^2 = 325 u = \pm\sqrt{325} = \pm \sqrt{325}
Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u
r =50 - \sqrt{325} = 31.972 s = 50 + \sqrt{325} = 68.028
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.