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\frac{50}{49}x^{2}-\frac{10}{49}x-\frac{24}{49}=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(-\frac{10}{49}\right)±\sqrt{\left(-\frac{10}{49}\right)^{2}-4\times \frac{50}{49}\left(-\frac{24}{49}\right)}}{2\times \frac{50}{49}}
This equation is in standard form: ax^{2}+bx+c=0. Substitute \frac{50}{49} for a, -\frac{10}{49} for b, and -\frac{24}{49} for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-\left(-\frac{10}{49}\right)±\sqrt{\frac{100}{2401}-4\times \frac{50}{49}\left(-\frac{24}{49}\right)}}{2\times \frac{50}{49}}
Square -\frac{10}{49} by squaring both the numerator and the denominator of the fraction.
x=\frac{-\left(-\frac{10}{49}\right)±\sqrt{\frac{100}{2401}-\frac{200}{49}\left(-\frac{24}{49}\right)}}{2\times \frac{50}{49}}
Multiply -4 times \frac{50}{49}.
x=\frac{-\left(-\frac{10}{49}\right)±\sqrt{\frac{100+4800}{2401}}}{2\times \frac{50}{49}}
Multiply -\frac{200}{49} times -\frac{24}{49} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.
x=\frac{-\left(-\frac{10}{49}\right)±\sqrt{\frac{100}{49}}}{2\times \frac{50}{49}}
Add \frac{100}{2401} to \frac{4800}{2401} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
x=\frac{-\left(-\frac{10}{49}\right)±\frac{10}{7}}{2\times \frac{50}{49}}
Take the square root of \frac{100}{49}.
x=\frac{\frac{10}{49}±\frac{10}{7}}{2\times \frac{50}{49}}
The opposite of -\frac{10}{49} is \frac{10}{49}.
x=\frac{\frac{10}{49}±\frac{10}{7}}{\frac{100}{49}}
Multiply 2 times \frac{50}{49}.
x=\frac{\frac{80}{49}}{\frac{100}{49}}
Now solve the equation x=\frac{\frac{10}{49}±\frac{10}{7}}{\frac{100}{49}} when ± is plus. Add \frac{10}{49} to \frac{10}{7} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
x=\frac{4}{5}
Divide \frac{80}{49} by \frac{100}{49} by multiplying \frac{80}{49} by the reciprocal of \frac{100}{49}.
x=-\frac{\frac{60}{49}}{\frac{100}{49}}
Now solve the equation x=\frac{\frac{10}{49}±\frac{10}{7}}{\frac{100}{49}} when ± is minus. Subtract \frac{10}{7} from \frac{10}{49} by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.
x=-\frac{3}{5}
Divide -\frac{60}{49} by \frac{100}{49} by multiplying -\frac{60}{49} by the reciprocal of \frac{100}{49}.
x=\frac{4}{5} x=-\frac{3}{5}
The equation is now solved.
\frac{50}{49}x^{2}-\frac{10}{49}x-\frac{24}{49}=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.
\frac{50}{49}x^{2}-\frac{10}{49}x-\frac{24}{49}-\left(-\frac{24}{49}\right)=-\left(-\frac{24}{49}\right)
Add \frac{24}{49} to both sides of the equation.
\frac{50}{49}x^{2}-\frac{10}{49}x=-\left(-\frac{24}{49}\right)
Subtracting -\frac{24}{49} from itself leaves 0.
\frac{50}{49}x^{2}-\frac{10}{49}x=\frac{24}{49}
Subtract -\frac{24}{49} from 0.
\frac{\frac{50}{49}x^{2}-\frac{10}{49}x}{\frac{50}{49}}=\frac{\frac{24}{49}}{\frac{50}{49}}
Divide both sides of the equation by \frac{50}{49}, which is the same as multiplying both sides by the reciprocal of the fraction.
x^{2}+\left(-\frac{\frac{10}{49}}{\frac{50}{49}}\right)x=\frac{\frac{24}{49}}{\frac{50}{49}}
Dividing by \frac{50}{49} undoes the multiplication by \frac{50}{49}.
x^{2}-\frac{1}{5}x=\frac{\frac{24}{49}}{\frac{50}{49}}
Divide -\frac{10}{49} by \frac{50}{49} by multiplying -\frac{10}{49} by the reciprocal of \frac{50}{49}.
x^{2}-\frac{1}{5}x=\frac{12}{25}
Divide \frac{24}{49} by \frac{50}{49} by multiplying \frac{24}{49} by the reciprocal of \frac{50}{49}.
x^{2}-\frac{1}{5}x+\left(-\frac{1}{10}\right)^{2}=\frac{12}{25}+\left(-\frac{1}{10}\right)^{2}
Divide -\frac{1}{5}, the coefficient of the x term, by 2 to get -\frac{1}{10}. Then add the square of -\frac{1}{10} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}-\frac{1}{5}x+\frac{1}{100}=\frac{12}{25}+\frac{1}{100}
Square -\frac{1}{10} by squaring both the numerator and the denominator of the fraction.
x^{2}-\frac{1}{5}x+\frac{1}{100}=\frac{49}{100}
Add \frac{12}{25} to \frac{1}{100} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
\left(x-\frac{1}{10}\right)^{2}=\frac{49}{100}
Factor x^{2}-\frac{1}{5}x+\frac{1}{100}. 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{1}{10}\right)^{2}}=\sqrt{\frac{49}{100}}
Take the square root of both sides of the equation.
x-\frac{1}{10}=\frac{7}{10} x-\frac{1}{10}=-\frac{7}{10}
Simplify.
x=\frac{4}{5} x=-\frac{3}{5}
Add \frac{1}{10} to both sides of the equation.