Solve 5x*(48-8x)=180

$$5 x \cdot ( 48 - 8 x ) = 180$$

Solve for x

$$x=\frac{3\sqrt{2}}{2}+3\approx 5.121320344$$

$$x=-\frac{3\sqrt{2}}{2}+3\approx 0.878679656$$

Steps Using the Quadratic Formula

Use the distributive property to multiply 5x by 48-8x.

$$240x-40x^{2}=180$$

Subtract 180 from both sides.

$$240x-40x^{2}-180=0$$

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.

$$-40x^{2}+240x-180=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -40 for a, 240 for b, and -180 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-240±\sqrt{240^{2}-4\left(-40\right)\left(-180\right)}}{2\left(-40\right)}$$

Square 240.

$$x=\frac{-240±\sqrt{57600-4\left(-40\right)\left(-180\right)}}{2\left(-40\right)}$$

Multiply -4 times -40.

$$x=\frac{-240±\sqrt{57600+160\left(-180\right)}}{2\left(-40\right)}$$

Multiply 160 times -180.

$$x=\frac{-240±\sqrt{57600-28800}}{2\left(-40\right)}$$

Add 57600 to -28800.

$$x=\frac{-240±\sqrt{28800}}{2\left(-40\right)}$$

Take the square root of 28800.

$$x=\frac{-240±120\sqrt{2}}{2\left(-40\right)}$$

Multiply 2 times -40.

$$x=\frac{-240±120\sqrt{2}}{-80}$$

Now solve the equation x=\frac{-240±120\sqrt{2}}{-80} when ± is plus. Add -240 to 120\sqrt{2}.

$$x=\frac{120\sqrt{2}-240}{-80}$$

Divide -240+120\sqrt{2} by -80.

$$x=-\frac{3\sqrt{2}}{2}+3$$

Now solve the equation x=\frac{-240±120\sqrt{2}}{-80} when ± is minus. Subtract 120\sqrt{2} from -240.

$$x=\frac{-120\sqrt{2}-240}{-80}$$

Divide -240-120\sqrt{2} by -80.

$$x=\frac{3\sqrt{2}}{2}+3$$

The equation is now solved.

$$x=-\frac{3\sqrt{2}}{2}+3 x=\frac{3\sqrt{2}}{2}+3$$

Steps for Completing the Square

Use the distributive property to multiply 5x by 48-8x.

$$240x-40x^{2}=180$$

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.

$$-40x^{2}+240x=180$$

Divide both sides by -40.

$$\frac{-40x^{2}+240x}{-40}=\frac{180}{-40}$$

Dividing by -40 undoes the multiplication by -40.

$$x^{2}+\frac{240}{-40}x=\frac{180}{-40}$$

Divide 240 by -40.

$$x^{2}-6x=\frac{180}{-40}$$

Reduce the fraction \frac{180}{-40} to lowest terms by extracting and canceling out 20.

$$x^{2}-6x=-\frac{9}{2}$$

Divide -6, the coefficient of the x term, by 2 to get -3. Then add the square of -3 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-6x+\left(-3\right)^{2}=-\frac{9}{2}+\left(-3\right)^{2}$$

Square -3.

$$x^{2}-6x+9=-\frac{9}{2}+9$$

Add -\frac{9}{2} to 9.

$$x^{2}-6x+9=\frac{9}{2}$$

Factor x^{2}-6x+9. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-3\right)^{2}=\frac{9}{2}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-3\right)^{2}}=\sqrt{\frac{9}{2}}$$

Simplify.

$$x-3=\frac{3\sqrt{2}}{2} x-3=-\frac{3\sqrt{2}}{2}$$

Add 3 to both sides of the equation.

$$x=\frac{3\sqrt{2}}{2}+3 x=-\frac{3\sqrt{2}}{2}+3$$