Solve 4x^2=12-3x

$$4 x ^ { 2 } = 12 - 3 x$$

Solve for x

$$x = \frac{\sqrt{201} - 3}{8} \approx 1.39718086$$

$$x=\frac{-\sqrt{201}-3}{8}\approx -2.14718086$$

Steps Using the Quadratic Formula

Subtract 12 from both sides.

$$4x^{2}-12=-3x$$

Add 3x to both sides.

$$4x^{2}-12+3x=0$$

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.

$$4x^{2}+3x-12=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 4 for a, 3 for b, and -12 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-3±\sqrt{3^{2}-4\times 4\left(-12\right)}}{2\times 4}$$

Square 3.

$$x=\frac{-3±\sqrt{9-4\times 4\left(-12\right)}}{2\times 4}$$

Multiply -4 times 4.

$$x=\frac{-3±\sqrt{9-16\left(-12\right)}}{2\times 4}$$

Multiply -16 times -12.

$$x=\frac{-3±\sqrt{9+192}}{2\times 4}$$

Add 9 to 192.

$$x=\frac{-3±\sqrt{201}}{2\times 4}$$

Multiply 2 times 4.

$$x=\frac{-3±\sqrt{201}}{8}$$

Now solve the equation x=\frac{-3±\sqrt{201}}{8} when ± is plus. Add -3 to \sqrt{201}.

$$x=\frac{\sqrt{201}-3}{8}$$

Now solve the equation x=\frac{-3±\sqrt{201}}{8} when ± is minus. Subtract \sqrt{201} from -3.

$$x=\frac{-\sqrt{201}-3}{8}$$

The equation is now solved.

$$x=\frac{\sqrt{201}-3}{8} x=\frac{-\sqrt{201}-3}{8}$$

Steps for Completing the Square

Add 3x to both sides.

$$4x^{2}+3x=12$$

Divide both sides by 4.

$$\frac{4x^{2}+3x}{4}=\frac{12}{4}$$

Dividing by 4 undoes the multiplication by 4.

$$x^{2}+\frac{3}{4}x=\frac{12}{4}$$

Divide 12 by 4.

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

Divide \frac{3}{4}, the coefficient of the x term, by 2 to get \frac{3}{8}. Then add the square of \frac{3}{8} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{3}{4}x+\left(\frac{3}{8}\right)^{2}=3+\left(\frac{3}{8}\right)^{2}$$

Square \frac{3}{8} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{3}{4}x+\frac{9}{64}=3+\frac{9}{64}$$

Add 3 to \frac{9}{64}.

$$x^{2}+\frac{3}{4}x+\frac{9}{64}=\frac{201}{64}$$

Factor x^{2}+\frac{3}{4}x+\frac{9}{64}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{3}{8}\right)^{2}=\frac{201}{64}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{3}{8}\right)^{2}}=\sqrt{\frac{201}{64}}$$

Simplify.

$$x+\frac{3}{8}=\frac{\sqrt{201}}{8} x+\frac{3}{8}=-\frac{\sqrt{201}}{8}$$

Subtract \frac{3}{8} from both sides of the equation.

$$x=\frac{\sqrt{201}-3}{8} x=\frac{-\sqrt{201}-3}{8}$$