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400x^{2}-144x-100=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(-144\right)±\sqrt{\left(-144\right)^{2}-4\times 400\left(-100\right)}}{2\times 400}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 400 for a, -144 for b, and -100 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-\left(-144\right)±\sqrt{20736-4\times 400\left(-100\right)}}{2\times 400}
Square -144.
x=\frac{-\left(-144\right)±\sqrt{20736-1600\left(-100\right)}}{2\times 400}
Multiply -4 times 400.
x=\frac{-\left(-144\right)±\sqrt{20736+160000}}{2\times 400}
Multiply -1600 times -100.
x=\frac{-\left(-144\right)±\sqrt{180736}}{2\times 400}
Add 20736 to 160000.
x=\frac{-\left(-144\right)±16\sqrt{706}}{2\times 400}
Take the square root of 180736.
x=\frac{144±16\sqrt{706}}{2\times 400}
The opposite of -144 is 144.
x=\frac{144±16\sqrt{706}}{800}
Multiply 2 times 400.
x=\frac{16\sqrt{706}+144}{800}
Now solve the equation x=\frac{144±16\sqrt{706}}{800} when ± is plus. Add 144 to 16\sqrt{706}.
x=\frac{\sqrt{706}+9}{50}
Divide 144+16\sqrt{706} by 800.
x=\frac{144-16\sqrt{706}}{800}
Now solve the equation x=\frac{144±16\sqrt{706}}{800} when ± is minus. Subtract 16\sqrt{706} from 144.
x=\frac{9-\sqrt{706}}{50}
Divide 144-16\sqrt{706} by 800.
x=\frac{\sqrt{706}+9}{50} x=\frac{9-\sqrt{706}}{50}
The equation is now solved.
400x^{2}-144x-100=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.
400x^{2}-144x-100-\left(-100\right)=-\left(-100\right)
Add 100 to both sides of the equation.
400x^{2}-144x=-\left(-100\right)
Subtracting -100 from itself leaves 0.
400x^{2}-144x=100
Subtract -100 from 0.
\frac{400x^{2}-144x}{400}=\frac{100}{400}
Divide both sides by 400.
x^{2}+\left(-\frac{144}{400}\right)x=\frac{100}{400}
Dividing by 400 undoes the multiplication by 400.
x^{2}-\frac{9}{25}x=\frac{100}{400}
Reduce the fraction \frac{-144}{400} to lowest terms by extracting and canceling out 16.
x^{2}-\frac{9}{25}x=\frac{1}{4}
Reduce the fraction \frac{100}{400} to lowest terms by extracting and canceling out 100.
x^{2}-\frac{9}{25}x+\left(-\frac{9}{50}\right)^{2}=\frac{1}{4}+\left(-\frac{9}{50}\right)^{2}
Divide -\frac{9}{25}, the coefficient of the x term, by 2 to get -\frac{9}{50}. Then add the square of -\frac{9}{50} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}-\frac{9}{25}x+\frac{81}{2500}=\frac{1}{4}+\frac{81}{2500}
Square -\frac{9}{50} by squaring both the numerator and the denominator of the fraction.
x^{2}-\frac{9}{25}x+\frac{81}{2500}=\frac{353}{1250}
Add \frac{1}{4} to \frac{81}{2500} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
\left(x-\frac{9}{50}\right)^{2}=\frac{353}{1250}
Factor x^{2}-\frac{9}{25}x+\frac{81}{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-\frac{9}{50}\right)^{2}}=\sqrt{\frac{353}{1250}}
Take the square root of both sides of the equation.
x-\frac{9}{50}=\frac{\sqrt{706}}{50} x-\frac{9}{50}=-\frac{\sqrt{706}}{50}
Simplify.
x=\frac{\sqrt{706}+9}{50} x=\frac{9-\sqrt{706}}{50}
Add \frac{9}{50} to both sides of the equation.
x ^ 2 -\frac{9}{25}x -\frac{1}{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 400
r + s = \frac{9}{25} rs = -\frac{1}{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{9}{50} - u s = \frac{9}{50} + u
Two numbers r and s sum up to \frac{9}{25} exactly when the average of the two numbers is \frac{1}{2}*\frac{9}{25} = \frac{9}{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>
(\frac{9}{50} - u) (\frac{9}{50} + u) = -\frac{1}{4}
To solve for unknown quantity u, substitute these in the product equation rs = -\frac{1}{4}
\frac{81}{2500} - u^2 = -\frac{1}{4}
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = -\frac{1}{4}-\frac{81}{2500} = -\frac{353}{1250}
Simplify the expression by subtracting \frac{81}{2500} on both sides
u^2 = \frac{353}{1250} u = \pm\sqrt{\frac{353}{1250}} = \pm \frac{\sqrt{353}}{\sqrt{1250}}
Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u
r =\frac{9}{50} - \frac{\sqrt{353}}{\sqrt{1250}} = -0.351 s = \frac{9}{50} + \frac{\sqrt{353}}{\sqrt{1250}} = 0.711
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.