Solve 6(1+x)+6*(1+x)(1+x)=17.34

$$6(1+x)+6 \times (1+x)(1+x)=17.34$$

Solve for x

$$x=\frac{\sqrt{314}}{10}-1.5\approx 0.272004515$$

$$x=-\frac{\sqrt{314}}{10}-1.5\approx -3.272004515$$

Steps Using the Quadratic Formula

Multiply 1+x and 1+x to get \left(1+x\right)^{2}.

$$6\left(1+x\right)+6\left(1+x\right)^{2}=17.34$$

Use the distributive property to multiply 6 by 1+x.

$$6+6x+6\left(1+x\right)^{2}=17.34$$

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

$$6+6x+6\left(1+2x+x^{2}\right)=17.34$$

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

$$6+6x+6+12x+6x^{2}=17.34$$

Add 6 and 6 to get 12.

$$12+6x+12x+6x^{2}=17.34$$

Combine 6x and 12x to get 18x.

$$12+18x+6x^{2}=17.34$$

Subtract 17.34 from both sides.

$$12+18x+6x^{2}-17.34=0$$

Subtract 17.34 from 12 to get -5.34.

$$-5.34+18x+6x^{2}=0$$

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.

$$6x^{2}+18x-5.34=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 6 for a, 18 for b, and -5.34 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-18±\sqrt{18^{2}-4\times 6\left(-5.34\right)}}{2\times 6}$$

Square 18.

$$x=\frac{-18±\sqrt{324-4\times 6\left(-5.34\right)}}{2\times 6}$$

Multiply -4 times 6.

$$x=\frac{-18±\sqrt{324-24\left(-5.34\right)}}{2\times 6}$$

Multiply -24 times -5.34.

$$x=\frac{-18±\sqrt{324+128.16}}{2\times 6}$$

Add 324 to 128.16.

$$x=\frac{-18±\sqrt{452.16}}{2\times 6}$$

Take the square root of 452.16.

$$x=\frac{-18±\frac{6\sqrt{314}}{5}}{2\times 6}$$

Multiply 2 times 6.

$$x=\frac{-18±\frac{6\sqrt{314}}{5}}{12}$$

Now solve the equation x=\frac{-18±\frac{6\sqrt{314}}{5}}{12} when ± is plus. Add -18 to \frac{6\sqrt{314}}{5}.

$$x=\frac{\frac{6\sqrt{314}}{5}-18}{12}$$

Divide -18+\frac{6\sqrt{314}}{5} by 12.

$$x=\frac{\sqrt{314}}{10}-\frac{3}{2}$$

Now solve the equation x=\frac{-18±\frac{6\sqrt{314}}{5}}{12} when ± is minus. Subtract \frac{6\sqrt{314}}{5} from -18.

$$x=\frac{-\frac{6\sqrt{314}}{5}-18}{12}$$

Divide -18-\frac{6\sqrt{314}}{5} by 12.

$$x=-\frac{\sqrt{314}}{10}-\frac{3}{2}$$

The equation is now solved.

$$x=\frac{\sqrt{314}}{10}-\frac{3}{2} x=-\frac{\sqrt{314}}{10}-\frac{3}{2}$$

Steps for Completing the Square

Multiply 1+x and 1+x to get \left(1+x\right)^{2}.

$$6\left(1+x\right)+6\left(1+x\right)^{2}=17.34$$

Use the distributive property to multiply 6 by 1+x.

$$6+6x+6\left(1+x\right)^{2}=17.34$$

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

$$6+6x+6\left(1+2x+x^{2}\right)=17.34$$

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

$$6+6x+6+12x+6x^{2}=17.34$$

Add 6 and 6 to get 12.

$$12+6x+12x+6x^{2}=17.34$$

Combine 6x and 12x to get 18x.

$$12+18x+6x^{2}=17.34$$

Subtract 12 from both sides.

$$18x+6x^{2}=17.34-12$$

Subtract 12 from 17.34 to get 5.34.

$$18x+6x^{2}=5.34$$

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.

$$6x^{2}+18x=5.34$$

Divide both sides by 6.

$$\frac{6x^{2}+18x}{6}=\frac{5.34}{6}$$

Dividing by 6 undoes the multiplication by 6.

$$x^{2}+\frac{18}{6}x=\frac{5.34}{6}$$

Divide 18 by 6.

$$x^{2}+3x=\frac{5.34}{6}$$

Divide 5.34 by 6.

$$x^{2}+3x=0.89$$

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.

$$x^{2}+3x+\left(\frac{3}{2}\right)^{2}=0.89+\left(\frac{3}{2}\right)^{2}$$

Square \frac{3}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+3x+\frac{9}{4}=0.89+\frac{9}{4}$$

Add 0.89 to \frac{9}{4} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+3x+\frac{9}{4}=\frac{157}{50}$$

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}.

$$\left(x+\frac{3}{2}\right)^{2}=\frac{157}{50}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{3}{2}\right)^{2}}=\sqrt{\frac{157}{50}}$$

Simplify.

$$x+\frac{3}{2}=\frac{\sqrt{314}}{10} x+\frac{3}{2}=-\frac{\sqrt{314}}{10}$$

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

$$x=\frac{\sqrt{314}}{10}-\frac{3}{2} x=-\frac{\sqrt{314}}{10}-\frac{3}{2}$$