Solve 6x=3-2x^2

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

Solve for x

$$x=\frac{\sqrt{15}-3}{2}\approx 0.436491673$$

$$x=\frac{-\sqrt{15}-3}{2}\approx -3.436491673$$

Steps Using the Quadratic Formula

Subtract 3 from both sides.

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

Add 2x^{2} to both sides.

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

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}.

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

Square 6.

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

Multiply -4 times 2.

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

Multiply -8 times -3.

$$x=\frac{-6±\sqrt{36+24}}{2\times 2}$$

Add 36 to 24.

$$x=\frac{-6±\sqrt{60}}{2\times 2}$$

Take the square root of 60.

$$x=\frac{-6±2\sqrt{15}}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{-6±2\sqrt{15}}{4}$$

Now solve the equation x=\frac{-6±2\sqrt{15}}{4} when ± is plus. Add -6 to 2\sqrt{15}.

$$x=\frac{2\sqrt{15}-6}{4}$$

Divide -6+2\sqrt{15} by 4.

$$x=\frac{\sqrt{15}-3}{2}$$

Now solve the equation x=\frac{-6±2\sqrt{15}}{4} when ± is minus. Subtract 2\sqrt{15} from -6.

$$x=\frac{-2\sqrt{15}-6}{4}$$

Divide -6-2\sqrt{15} by 4.

$$x=\frac{-\sqrt{15}-3}{2}$$

The equation is now solved.

$$x=\frac{\sqrt{15}-3}{2} x=\frac{-\sqrt{15}-3}{2}$$

Steps for Completing the Square

Add 2x^{2} to both sides.

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

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, 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.

$$x^{2}+3x+\frac{9}{4}=\frac{15}{4}$$

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}.

$$\left(x+\frac{3}{2}\right)^{2}=\frac{15}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{3}{2}\right)^{2}}=\sqrt{\frac{15}{4}}$$

Simplify.

$$x+\frac{3}{2}=\frac{\sqrt{15}}{2} x+\frac{3}{2}=-\frac{\sqrt{15}}{2}$$

Subtract \frac{3}{2} from both sides of the equation.

$$x=\frac{\sqrt{15}-3}{2} x=\frac{-\sqrt{15}-3}{2}$$