Solve x*6-2x/2=1.5

$$x \cdot \frac { 6 - 2 x } { 2 } = 1.5$$

Solve for x

$$x = \frac{\sqrt{3} + 3}{2} \approx 2.366025404$$

$$x=\frac{3-\sqrt{3}}{2}\approx 0.633974596$$

Steps Using the Quadratic Formula

Multiply both sides of the equation by 2.

$$x\left(6-2x\right)=3$$

Use the distributive property to multiply x by 6-2x.

$$6x-2x^{2}=3$$

Subtract 3 from both sides.

$$6x-2x^{2}-3=0$$

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.

$$-2x^{2}+6x-3=0$$

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

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

Square 6.

$$x=\frac{-6±\sqrt{36-4\left(-2\right)\left(-3\right)}}{2\left(-2\right)}$$

Multiply -4 times -2.

$$x=\frac{-6±\sqrt{36+8\left(-3\right)}}{2\left(-2\right)}$$

Multiply 8 times -3.

$$x=\frac{-6±\sqrt{36-24}}{2\left(-2\right)}$$

Add 36 to -24.

$$x=\frac{-6±\sqrt{12}}{2\left(-2\right)}$$

Take the square root of 12.

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

Multiply 2 times -2.

$$x=\frac{-6±2\sqrt{3}}{-4}$$

Now solve the equation x=\frac{-6±2\sqrt{3}}{-4} when ± is plus. Add -6 to 2\sqrt{3}.

$$x=\frac{2\sqrt{3}-6}{-4}$$

Divide -6+2\sqrt{3} by -4.

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

Now solve the equation x=\frac{-6±2\sqrt{3}}{-4} when ± is minus. Subtract 2\sqrt{3} from -6.

$$x=\frac{-2\sqrt{3}-6}{-4}$$

Divide -6-2\sqrt{3} by -4.

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

The equation is now solved.

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

Steps for Completing the Square

Multiply both sides of the equation by 2.

$$x\left(6-2x\right)=3$$

Use the distributive property to multiply x by 6-2x.

$$6x-2x^{2}=3$$

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.

$$-2x^{2}+6x=3$$

Divide both sides by -2.

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

Dividing by -2 undoes the multiplication by -2.

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

Divide 6 by -2.

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

Divide 3 by -2.

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

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

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

Square -\frac{3}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-3x+\frac{9}{4}=-\frac{3}{2}+\frac{9}{4}$$

Add -\frac{3}{2} to \frac{9}{4} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-3x+\frac{9}{4}=\frac{3}{4}$$

Factor x^{2}-3x+\frac{9}{4}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

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

Take the square root of both sides of the equation.

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

Simplify.

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

Add \frac{3}{2} to both sides of the equation.

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