Solve 130x^2-540x+560=0

$$130 { x }^{ 2 } -540x+560 = 0$$

Solve for x

$$x=2$$

$$x = \frac{28}{13} = 2\frac{2}{13} \approx 2.153846154$$

Steps Using the Quadratic Formula

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.

$$130x^{2}-540x+560=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 130 for a, -540 for b, and 560 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-540\right)±\sqrt{\left(-540\right)^{2}-4\times 130\times 560}}{2\times 130}$$

Square -540.

$$x=\frac{-\left(-540\right)±\sqrt{291600-4\times 130\times 560}}{2\times 130}$$

Multiply -4 times 130.

$$x=\frac{-\left(-540\right)±\sqrt{291600-520\times 560}}{2\times 130}$$

Multiply -520 times 560.

$$x=\frac{-\left(-540\right)±\sqrt{291600-291200}}{2\times 130}$$

Add 291600 to -291200.

$$x=\frac{-\left(-540\right)±\sqrt{400}}{2\times 130}$$

Take the square root of 400.

$$x=\frac{-\left(-540\right)±20}{2\times 130}$$

The opposite of -540 is 540.

$$x=\frac{540±20}{2\times 130}$$

Multiply 2 times 130.

$$x=\frac{540±20}{260}$$

Now solve the equation x=\frac{540±20}{260} when ± is plus. Add 540 to 20.

$$x=\frac{560}{260}$$

Reduce the fraction \frac{560}{260} to lowest terms by extracting and canceling out 20.

$$x=\frac{28}{13}$$

Now solve the equation x=\frac{540±20}{260} when ± is minus. Subtract 20 from 540.

$$x=\frac{520}{260}$$

Divide 520 by 260.

$$x=2$$

The equation is now solved.

$$x=\frac{28}{13} x=2$$

Steps for Completing the Square

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.

$$130x^{2}-540x+560=0$$

Subtract 560 from both sides of the equation.

$$130x^{2}-540x+560-560=-560$$

Subtracting 560 from itself leaves 0.

$$130x^{2}-540x=-560$$

Divide both sides by 130.

$$\frac{130x^{2}-540x}{130}=-\frac{560}{130}$$

Dividing by 130 undoes the multiplication by 130.

$$x^{2}+\left(-\frac{540}{130}\right)x=-\frac{560}{130}$$

Reduce the fraction \frac{-540}{130} to lowest terms by extracting and canceling out 10.

$$x^{2}-\frac{54}{13}x=-\frac{560}{130}$$

Reduce the fraction \frac{-560}{130} to lowest terms by extracting and canceling out 10.

$$x^{2}-\frac{54}{13}x=-\frac{56}{13}$$

Divide -\frac{54}{13}, the coefficient of the x term, by 2 to get -\frac{27}{13}. Then add the square of -\frac{27}{13} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{54}{13}x+\left(-\frac{27}{13}\right)^{2}=-\frac{56}{13}+\left(-\frac{27}{13}\right)^{2}$$

Square -\frac{27}{13} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{54}{13}x+\frac{729}{169}=-\frac{56}{13}+\frac{729}{169}$$

Add -\frac{56}{13} to \frac{729}{169} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{54}{13}x+\frac{729}{169}=\frac{1}{169}$$

Factor x^{2}-\frac{54}{13}x+\frac{729}{169}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{27}{13}\right)^{2}=\frac{1}{169}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{27}{13}\right)^{2}}=\sqrt{\frac{1}{169}}$$

Simplify.

$$x-\frac{27}{13}=\frac{1}{13} x-\frac{27}{13}=-\frac{1}{13}$$

Add \frac{27}{13} to both sides of the equation.

$$x=\frac{28}{13} x=2$$