Solve (2)4x^2-6x-3=0

$$( 2 ) 4 x ^ { 2 } - 6 x - 3 = 0$$

Solve for x

$$x = \frac{\sqrt{33} + 3}{8} \approx 1.093070331$$

$$x=\frac{3-\sqrt{33}}{8}\approx -0.343070331$$

Steps Using the Quadratic Formula

Multiply 2 and 4 to get 8.

$$8x^{2}-6x-3=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 8 for a, -6 for b, and -3 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-6\right)±\sqrt{\left(-6\right)^{2}-4\times 8\left(-3\right)}}{2\times 8}$$

Square -6.

$$x=\frac{-\left(-6\right)±\sqrt{36-4\times 8\left(-3\right)}}{2\times 8}$$

Multiply -4 times 8.

$$x=\frac{-\left(-6\right)±\sqrt{36-32\left(-3\right)}}{2\times 8}$$

Multiply -32 times -3.

$$x=\frac{-\left(-6\right)±\sqrt{36+96}}{2\times 8}$$

Add 36 to 96.

$$x=\frac{-\left(-6\right)±\sqrt{132}}{2\times 8}$$

Take the square root of 132.

$$x=\frac{-\left(-6\right)±2\sqrt{33}}{2\times 8}$$

The opposite of -6 is 6.

$$x=\frac{6±2\sqrt{33}}{2\times 8}$$

Multiply 2 times 8.

$$x=\frac{6±2\sqrt{33}}{16}$$

Now solve the equation x=\frac{6±2\sqrt{33}}{16} when ± is plus. Add 6 to 2\sqrt{33}.

$$x=\frac{2\sqrt{33}+6}{16}$$

Divide 6+2\sqrt{33} by 16.

$$x=\frac{\sqrt{33}+3}{8}$$

Now solve the equation x=\frac{6±2\sqrt{33}}{16} when ± is minus. Subtract 2\sqrt{33} from 6.

$$x=\frac{6-2\sqrt{33}}{16}$$

Divide 6-2\sqrt{33} by 16.

$$x=\frac{3-\sqrt{33}}{8}$$

The equation is now solved.

$$x=\frac{\sqrt{33}+3}{8} x=\frac{3-\sqrt{33}}{8}$$

Steps for Completing the Square

Multiply 2 and 4 to get 8.

$$8x^{2}-6x-3=0$$

Add 3 to both sides. Anything plus zero gives itself.

$$8x^{2}-6x=3$$

Divide both sides by 8.

$$\frac{8x^{2}-6x}{8}=\frac{3}{8}$$

Dividing by 8 undoes the multiplication by 8.

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

Reduce the fraction \frac{-6}{8} to lowest terms by extracting and canceling out 2.

$$x^{2}-\frac{3}{4}x=\frac{3}{8}$$

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.

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

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

$$x^{2}-\frac{3}{4}x+\frac{9}{64}=\frac{3}{8}+\frac{9}{64}$$

Add \frac{3}{8} to \frac{9}{64} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{3}{4}x+\frac{9}{64}=\frac{33}{64}$$

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}.

$$\left(x-\frac{3}{8}\right)^{2}=\frac{33}{64}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{3}{8}\right)^{2}}=\sqrt{\frac{33}{64}}$$

Simplify.

$$x-\frac{3}{8}=\frac{\sqrt{33}}{8} x-\frac{3}{8}=-\frac{\sqrt{33}}{8}$$

Add \frac{3}{8} to both sides of the equation.

$$x=\frac{\sqrt{33}+3}{8} x=\frac{3-\sqrt{33}}{8}$$