Solve 2(3x+1)^2/5=8

$$\frac { 2 ( 3 x + 1 ) ^ { 2 } } { 5 } = 8$$

Solve for x

$$x = \frac{2 \sqrt{5} - 1}{3} \approx 1.157378652$$

$$x=\frac{-2\sqrt{5}-1}{3}\approx -1.824045318$$

Steps Using the Quadratic Formula

Multiply both sides by 5.

$$2\left(3x+1\right)^{2}=8\times 5$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(3x+1\right)^{2}.

$$2\left(9x^{2}+6x+1\right)=8\times 5$$

Use the distributive property to multiply 2 by 9x^{2}+6x+1.

$$18x^{2}+12x+2=8\times 5$$

Multiply 8 and 5 to get 40.

$$18x^{2}+12x+2=40$$

Subtract 40 from both sides.

$$18x^{2}+12x+2-40=0$$

Subtract 40 from 2 to get -38.

$$18x^{2}+12x-38=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 18 for a, 12 for b, and -38 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-12±\sqrt{12^{2}-4\times 18\left(-38\right)}}{2\times 18}$$

Square 12.

$$x=\frac{-12±\sqrt{144-4\times 18\left(-38\right)}}{2\times 18}$$

Multiply -4 times 18.

$$x=\frac{-12±\sqrt{144-72\left(-38\right)}}{2\times 18}$$

Multiply -72 times -38.

$$x=\frac{-12±\sqrt{144+2736}}{2\times 18}$$

Add 144 to 2736.

$$x=\frac{-12±\sqrt{2880}}{2\times 18}$$

Take the square root of 2880.

$$x=\frac{-12±24\sqrt{5}}{2\times 18}$$

Multiply 2 times 18.

$$x=\frac{-12±24\sqrt{5}}{36}$$

Now solve the equation x=\frac{-12±24\sqrt{5}}{36} when ± is plus. Add -12 to 24\sqrt{5}.

$$x=\frac{24\sqrt{5}-12}{36}$$

Divide -12+24\sqrt{5} by 36.

$$x=\frac{2\sqrt{5}-1}{3}$$

Now solve the equation x=\frac{-12±24\sqrt{5}}{36} when ± is minus. Subtract 24\sqrt{5} from -12.

$$x=\frac{-24\sqrt{5}-12}{36}$$

Divide -12-24\sqrt{5} by 36.

$$x=\frac{-2\sqrt{5}-1}{3}$$

The equation is now solved.

$$x=\frac{2\sqrt{5}-1}{3} x=\frac{-2\sqrt{5}-1}{3}$$

Steps for Completing the Square

Multiply both sides by 5.

$$2\left(3x+1\right)^{2}=8\times 5$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(3x+1\right)^{2}.

$$2\left(9x^{2}+6x+1\right)=8\times 5$$

Use the distributive property to multiply 2 by 9x^{2}+6x+1.

$$18x^{2}+12x+2=8\times 5$$

Multiply 8 and 5 to get 40.

$$18x^{2}+12x+2=40$$

Subtract 2 from both sides.

$$18x^{2}+12x=40-2$$

Subtract 2 from 40 to get 38.

$$18x^{2}+12x=38$$

Divide both sides by 18.

$$\frac{18x^{2}+12x}{18}=\frac{38}{18}$$

Dividing by 18 undoes the multiplication by 18.

$$x^{2}+\frac{12}{18}x=\frac{38}{18}$$

Reduce the fraction \frac{12}{18} to lowest terms by extracting and canceling out 6.

$$x^{2}+\frac{2}{3}x=\frac{38}{18}$$

Reduce the fraction \frac{38}{18} to lowest terms by extracting and canceling out 2.

$$x^{2}+\frac{2}{3}x=\frac{19}{9}$$

Divide \frac{2}{3}, the coefficient of the x term, by 2 to get \frac{1}{3}. Then add the square of \frac{1}{3} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{2}{3}x+\left(\frac{1}{3}\right)^{2}=\frac{19}{9}+\left(\frac{1}{3}\right)^{2}$$

Square \frac{1}{3} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{2}{3}x+\frac{1}{9}=\frac{19+1}{9}$$

Add \frac{19}{9} to \frac{1}{9} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{2}{3}x+\frac{1}{9}=\frac{20}{9}$$

Factor x^{2}+\frac{2}{3}x+\frac{1}{9}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{1}{3}\right)^{2}=\frac{20}{9}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{1}{3}\right)^{2}}=\sqrt{\frac{20}{9}}$$

Simplify.

$$x+\frac{1}{3}=\frac{2\sqrt{5}}{3} x+\frac{1}{3}=-\frac{2\sqrt{5}}{3}$$

Subtract \frac{1}{3} from both sides of the equation.

$$x=\frac{2\sqrt{5}-1}{3} x=\frac{-2\sqrt{5}-1}{3}$$