Solve 160x^2-72x+4.5=0

$$160 x ^ { 2 } - 72 x + 4.5 = 0$$

Solve for x

$$x=0.075$$

$$x=0.375$$

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.

$$160x^{2}-72x+4.5=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 160 for a, -72 for b, and 4.5 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-72\right)±\sqrt{\left(-72\right)^{2}-4\times 160\times 4.5}}{2\times 160}$$

Square -72.

$$x=\frac{-\left(-72\right)±\sqrt{5184-4\times 160\times 4.5}}{2\times 160}$$

Multiply -4 times 160.

$$x=\frac{-\left(-72\right)±\sqrt{5184-640\times 4.5}}{2\times 160}$$

Multiply -640 times 4.5.

$$x=\frac{-\left(-72\right)±\sqrt{5184-2880}}{2\times 160}$$

Add 5184 to -2880.

$$x=\frac{-\left(-72\right)±\sqrt{2304}}{2\times 160}$$

Take the square root of 2304.

$$x=\frac{-\left(-72\right)±48}{2\times 160}$$

The opposite of -72 is 72.

$$x=\frac{72±48}{2\times 160}$$

Multiply 2 times 160.

$$x=\frac{72±48}{320}$$

Now solve the equation x=\frac{72±48}{320} when ± is plus. Add 72 to 48.

$$x=\frac{120}{320}$$

Reduce the fraction \frac{120}{320} to lowest terms by extracting and canceling out 40.

$$x=\frac{3}{8}$$

Now solve the equation x=\frac{72±48}{320} when ± is minus. Subtract 48 from 72.

$$x=\frac{24}{320}$$

Reduce the fraction \frac{24}{320} to lowest terms by extracting and canceling out 8.

$$x=\frac{3}{40}$$

The equation is now solved.

$$x=\frac{3}{8} x=\frac{3}{40}$$

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.

$$160x^{2}-72x+4.5=0$$

Subtract 4.5 from both sides of the equation.

$$160x^{2}-72x+4.5-4.5=-4.5$$

Subtracting 4.5 from itself leaves 0.

$$160x^{2}-72x=-4.5$$

Divide both sides by 160.

$$\frac{160x^{2}-72x}{160}=-\frac{4.5}{160}$$

Dividing by 160 undoes the multiplication by 160.

$$x^{2}+\left(-\frac{72}{160}\right)x=-\frac{4.5}{160}$$

Reduce the fraction \frac{-72}{160} to lowest terms by extracting and canceling out 8.

$$x^{2}-\frac{9}{20}x=-\frac{4.5}{160}$$

Divide -4.5 by 160.

$$x^{2}-\frac{9}{20}x=-0.028125$$

Divide -\frac{9}{20}, the coefficient of the x term, by 2 to get -\frac{9}{40}. Then add the square of -\frac{9}{40} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{9}{20}x+\left(-\frac{9}{40}\right)^{2}=-0.028125+\left(-\frac{9}{40}\right)^{2}$$

Square -\frac{9}{40} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{9}{20}x+\frac{81}{1600}=-0.028125+\frac{81}{1600}$$

Add -0.028125 to \frac{81}{1600} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{9}{20}x+\frac{81}{1600}=\frac{9}{400}$$

Factor x^{2}-\frac{9}{20}x+\frac{81}{1600}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{9}{40}\right)^{2}=\frac{9}{400}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{9}{40}\right)^{2}}=\sqrt{\frac{9}{400}}$$

Simplify.

$$x-\frac{9}{40}=\frac{3}{20} x-\frac{9}{40}=-\frac{3}{20}$$

Add \frac{9}{40} to both sides of the equation.

$$x=\frac{3}{8} x=\frac{3}{40}$$