Solve (x)2(2x+1)=3(x+4)

$$( x ) 2 ( 2 x + 1 ) = 3 ( x + 4 )$$

Solve for x

$$x = \frac{\sqrt{193} + 1}{8} \approx 1.861555499$$

$$x=\frac{1-\sqrt{193}}{8}\approx -1.611555499$$

Steps Using the Quadratic Formula

Use the distributive property to multiply x\times 2 by 2x+1.

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

Use the distributive property to multiply 3 by x+4.

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

Subtract 3x from both sides.

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

Combine x\times 2 and -3x to get -x.

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

Subtract 12 from both sides.

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

This equation is in standard form: ax^{2}+bx+c=0. Substitute 4 for a, -1 for b, and -12 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-1\right)±\sqrt{1-4\times 4\left(-12\right)}}{2\times 4}$$

Multiply -4 times 4.

$$x=\frac{-\left(-1\right)±\sqrt{1-16\left(-12\right)}}{2\times 4}$$

Multiply -16 times -12.

$$x=\frac{-\left(-1\right)±\sqrt{1+192}}{2\times 4}$$

Add 1 to 192.

$$x=\frac{-\left(-1\right)±\sqrt{193}}{2\times 4}$$

The opposite of -1 is 1.

$$x=\frac{1±\sqrt{193}}{2\times 4}$$

Multiply 2 times 4.

$$x=\frac{1±\sqrt{193}}{8}$$

Now solve the equation x=\frac{1±\sqrt{193}}{8} when ± is plus. Add 1 to \sqrt{193}.

$$x=\frac{\sqrt{193}+1}{8}$$

Now solve the equation x=\frac{1±\sqrt{193}}{8} when ± is minus. Subtract \sqrt{193} from 1.

$$x=\frac{1-\sqrt{193}}{8}$$

The equation is now solved.

$$x=\frac{\sqrt{193}+1}{8} x=\frac{1-\sqrt{193}}{8}$$

Steps for Completing the Square

Use the distributive property to multiply x\times 2 by 2x+1.

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

Use the distributive property to multiply 3 by x+4.

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

Subtract 3x from both sides.

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

Combine x\times 2 and -3x to get -x.

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

Divide both sides by 4.

$$\frac{4x^{2}-x}{4}=\frac{12}{4}$$

Dividing by 4 undoes the multiplication by 4.

$$x^{2}-\frac{1}{4}x=\frac{12}{4}$$

Divide 12 by 4.

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

Divide -\frac{1}{4}, the coefficient of the x term, by 2 to get -\frac{1}{8}. Then add the square of -\frac{1}{8} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{1}{4}x+\left(-\frac{1}{8}\right)^{2}=3+\left(-\frac{1}{8}\right)^{2}$$

Square -\frac{1}{8} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{1}{4}x+\frac{1}{64}=3+\frac{1}{64}$$

Add 3 to \frac{1}{64}.

$$x^{2}-\frac{1}{4}x+\frac{1}{64}=\frac{193}{64}$$

Factor x^{2}-\frac{1}{4}x+\frac{1}{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}.

$$\left(x-\frac{1}{8}\right)^{2}=\frac{193}{64}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{1}{8}\right)^{2}}=\sqrt{\frac{193}{64}}$$

Simplify.

$$x-\frac{1}{8}=\frac{\sqrt{193}}{8} x-\frac{1}{8}=-\frac{\sqrt{193}}{8}$$

Add \frac{1}{8} to both sides of the equation.

$$x=\frac{\sqrt{193}+1}{8} x=\frac{1-\sqrt{193}}{8}$$