Solve 490x^2-1120x+240=0

$$490 { x }^{ 2 } -1120x+240=0$$

Solve for x

$$x = \frac{2 \sqrt{10} + 8}{7} \approx 2.046365046$$

$$x=\frac{8-2\sqrt{10}}{7}\approx 0.23934924$$

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.

$$490x^{2}-1120x+240=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 490 for a, -1120 for b, and 240 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-1120\right)±\sqrt{\left(-1120\right)^{2}-4\times 490\times 240}}{2\times 490}$$

Square -1120.

$$x=\frac{-\left(-1120\right)±\sqrt{1254400-4\times 490\times 240}}{2\times 490}$$

Multiply -4 times 490.

$$x=\frac{-\left(-1120\right)±\sqrt{1254400-1960\times 240}}{2\times 490}$$

Multiply -1960 times 240.

$$x=\frac{-\left(-1120\right)±\sqrt{1254400-470400}}{2\times 490}$$

Add 1254400 to -470400.

$$x=\frac{-\left(-1120\right)±\sqrt{784000}}{2\times 490}$$

Take the square root of 784000.

$$x=\frac{-\left(-1120\right)±280\sqrt{10}}{2\times 490}$$

The opposite of -1120 is 1120.

$$x=\frac{1120±280\sqrt{10}}{2\times 490}$$

Multiply 2 times 490.

$$x=\frac{1120±280\sqrt{10}}{980}$$

Now solve the equation x=\frac{1120±280\sqrt{10}}{980} when ± is plus. Add 1120 to 280\sqrt{10}.

$$x=\frac{280\sqrt{10}+1120}{980}$$

Divide 1120+280\sqrt{10} by 980.

$$x=\frac{2\sqrt{10}+8}{7}$$

Now solve the equation x=\frac{1120±280\sqrt{10}}{980} when ± is minus. Subtract 280\sqrt{10} from 1120.

$$x=\frac{1120-280\sqrt{10}}{980}$$

Divide 1120-280\sqrt{10} by 980.

$$x=\frac{8-2\sqrt{10}}{7}$$

The equation is now solved.

$$x=\frac{2\sqrt{10}+8}{7} x=\frac{8-2\sqrt{10}}{7}$$

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.

$$490x^{2}-1120x+240=0$$

Subtract 240 from both sides of the equation.

$$490x^{2}-1120x+240-240=-240$$

Subtracting 240 from itself leaves 0.

$$490x^{2}-1120x=-240$$

Divide both sides by 490.

$$\frac{490x^{2}-1120x}{490}=-\frac{240}{490}$$

Dividing by 490 undoes the multiplication by 490.

$$x^{2}+\left(-\frac{1120}{490}\right)x=-\frac{240}{490}$$

Reduce the fraction \frac{-1120}{490} to lowest terms by extracting and canceling out 70.

$$x^{2}-\frac{16}{7}x=-\frac{240}{490}$$

Reduce the fraction \frac{-240}{490} to lowest terms by extracting and canceling out 10.

$$x^{2}-\frac{16}{7}x=-\frac{24}{49}$$

Divide -\frac{16}{7}, the coefficient of the x term, by 2 to get -\frac{8}{7}. Then add the square of -\frac{8}{7} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{16}{7}x+\left(-\frac{8}{7}\right)^{2}=-\frac{24}{49}+\left(-\frac{8}{7}\right)^{2}$$

Square -\frac{8}{7} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{16}{7}x+\frac{64}{49}=\frac{-24+64}{49}$$

Add -\frac{24}{49} to \frac{64}{49} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{16}{7}x+\frac{64}{49}=\frac{40}{49}$$

Factor x^{2}-\frac{16}{7}x+\frac{64}{49}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{8}{7}\right)^{2}=\frac{40}{49}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{8}{7}\right)^{2}}=\sqrt{\frac{40}{49}}$$

Simplify.

$$x-\frac{8}{7}=\frac{2\sqrt{10}}{7} x-\frac{8}{7}=-\frac{2\sqrt{10}}{7}$$

Add \frac{8}{7} to both sides of the equation.

$$x=\frac{2\sqrt{10}+8}{7} x=\frac{8-2\sqrt{10}}{7}$$