Solve 1/2x(x+1.5)=6

$$\frac { 1 } { 2 } x ( x + 1.5 ) = 6$$

Solve for x

$$x = \frac{\sqrt{201} - 3}{4} \approx 2.79436172$$

$$x=\frac{-\sqrt{201}-3}{4}\approx -4.29436172$$

Steps Using the Quadratic Formula

Use the distributive property to multiply \frac{1}{2}x by x+1.5.

$$\frac{1}{2}xx+\frac{1}{2}x\times 1.5=6$$

Multiply x and x to get x^{2}.

$$\frac{1}{2}x^{2}+\frac{1}{2}x\times 1.5=6$$

Convert decimal number 1.5 to fraction \frac{15}{10}. Reduce the fraction \frac{15}{10} to lowest terms by extracting and canceling out 5.

$$\frac{1}{2}x^{2}+\frac{1}{2}x\times \frac{3}{2}=6$$

Multiply \frac{1}{2} times \frac{3}{2} by multiplying numerator times numerator and denominator times denominator.

$$\frac{1}{2}x^{2}+\frac{1\times 3}{2\times 2}x=6$$

Do the multiplications in the fraction \frac{1\times 3}{2\times 2}.

$$\frac{1}{2}x^{2}+\frac{3}{4}x=6$$

Subtract 6 from both sides.

$$\frac{1}{2}x^{2}+\frac{3}{4}x-6=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute \frac{1}{2} for a, \frac{3}{4} for b, and -6 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\frac{3}{4}±\sqrt{\left(\frac{3}{4}\right)^{2}-4\times \frac{1}{2}\left(-6\right)}}{2\times \frac{1}{2}}$$

Square \frac{3}{4} by squaring both the numerator and the denominator of the fraction.

$$x=\frac{-\frac{3}{4}±\sqrt{\frac{9}{16}-4\times \frac{1}{2}\left(-6\right)}}{2\times \frac{1}{2}}$$

Multiply -4 times \frac{1}{2}.

$$x=\frac{-\frac{3}{4}±\sqrt{\frac{9}{16}-2\left(-6\right)}}{2\times \frac{1}{2}}$$

Multiply -2 times -6.

$$x=\frac{-\frac{3}{4}±\sqrt{\frac{9}{16}+12}}{2\times \frac{1}{2}}$$

Add \frac{9}{16} to 12.

$$x=\frac{-\frac{3}{4}±\sqrt{\frac{201}{16}}}{2\times \frac{1}{2}}$$

Take the square root of \frac{201}{16}.

$$x=\frac{-\frac{3}{4}±\frac{\sqrt{201}}{4}}{2\times \frac{1}{2}}$$

Multiply 2 times \frac{1}{2}.

$$x=\frac{-\frac{3}{4}±\frac{\sqrt{201}}{4}}{1}$$

Now solve the equation x=\frac{-\frac{3}{4}±\frac{\sqrt{201}}{4}}{1} when ± is plus. Add -\frac{3}{4} to \frac{\sqrt{201}}{4}.

$$x=\frac{\sqrt{201}-3}{4}$$

Now solve the equation x=\frac{-\frac{3}{4}±\frac{\sqrt{201}}{4}}{1} when ± is minus. Subtract \frac{\sqrt{201}}{4} from -\frac{3}{4}.

$$x=\frac{-\sqrt{201}-3}{4}$$

The equation is now solved.

$$x=\frac{\sqrt{201}-3}{4} x=\frac{-\sqrt{201}-3}{4}$$

Steps for Completing the Square

Use the distributive property to multiply \frac{1}{2}x by x+1.5.

$$\frac{1}{2}xx+\frac{1}{2}x\times 1.5=6$$

Multiply x and x to get x^{2}.

$$\frac{1}{2}x^{2}+\frac{1}{2}x\times 1.5=6$$

Convert decimal number 1.5 to fraction \frac{15}{10}. Reduce the fraction \frac{15}{10} to lowest terms by extracting and canceling out 5.

$$\frac{1}{2}x^{2}+\frac{1}{2}x\times \frac{3}{2}=6$$

Multiply \frac{1}{2} times \frac{3}{2} by multiplying numerator times numerator and denominator times denominator.

$$\frac{1}{2}x^{2}+\frac{1\times 3}{2\times 2}x=6$$

Do the multiplications in the fraction \frac{1\times 3}{2\times 2}.

$$\frac{1}{2}x^{2}+\frac{3}{4}x=6$$

Multiply both sides by 2.

$$\frac{\frac{1}{2}x^{2}+\frac{3}{4}x}{\frac{1}{2}}=\frac{6}{\frac{1}{2}}$$

Dividing by \frac{1}{2} undoes the multiplication by \frac{1}{2}.

$$x^{2}+\frac{\frac{3}{4}}{\frac{1}{2}}x=\frac{6}{\frac{1}{2}}$$

Divide \frac{3}{4} by \frac{1}{2} by multiplying \frac{3}{4} by the reciprocal of \frac{1}{2}.

$$x^{2}+\frac{3}{2}x=\frac{6}{\frac{1}{2}}$$

Divide 6 by \frac{1}{2} by multiplying 6 by the reciprocal of \frac{1}{2}.

$$x^{2}+\frac{3}{2}x=12$$

Divide \frac{3}{2}, the coefficient of the x term, by 2 to get \frac{3}{4}. Then add the square of \frac{3}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{3}{2}x+\left(\frac{3}{4}\right)^{2}=12+\left(\frac{3}{4}\right)^{2}$$

Square \frac{3}{4} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{3}{2}x+\frac{9}{16}=12+\frac{9}{16}$$

Add 12 to \frac{9}{16}.

$$x^{2}+\frac{3}{2}x+\frac{9}{16}=\frac{201}{16}$$

Factor x^{2}+\frac{3}{2}x+\frac{9}{16}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{3}{4}\right)^{2}=\frac{201}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{3}{4}\right)^{2}}=\sqrt{\frac{201}{16}}$$

Simplify.

$$x+\frac{3}{4}=\frac{\sqrt{201}}{4} x+\frac{3}{4}=-\frac{\sqrt{201}}{4}$$

Subtract \frac{3}{4} from both sides of the equation.

$$x=\frac{\sqrt{201}-3}{4} x=\frac{-\sqrt{201}-3}{4}$$