Solve 6x+3x(3x+1)+9=90

$$6x+3x(3x+1)+9 = 90$$

Solve for x

$$x = \frac{\sqrt{37} - 1}{2} \approx 2.541381265$$

$$x=\frac{-\sqrt{37}-1}{2}\approx -3.541381265$$

Steps Using the Quadratic Formula

Use the distributive property to multiply 3x by 3x+1.

$$6x+9x^{2}+3x+9=90$$

Combine 6x and 3x to get 9x.

$$9x+9x^{2}+9=90$$

Subtract 90 from both sides.

$$9x+9x^{2}+9-90=0$$

Subtract 90 from 9 to get -81.

$$9x+9x^{2}-81=0$$

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.

$$9x^{2}+9x-81=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 9 for a, 9 for b, and -81 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-9±\sqrt{9^{2}-4\times 9\left(-81\right)}}{2\times 9}$$

Square 9.

$$x=\frac{-9±\sqrt{81-4\times 9\left(-81\right)}}{2\times 9}$$

Multiply -4 times 9.

$$x=\frac{-9±\sqrt{81-36\left(-81\right)}}{2\times 9}$$

Multiply -36 times -81.

$$x=\frac{-9±\sqrt{81+2916}}{2\times 9}$$

Add 81 to 2916.

$$x=\frac{-9±\sqrt{2997}}{2\times 9}$$

Take the square root of 2997.

$$x=\frac{-9±9\sqrt{37}}{2\times 9}$$

Multiply 2 times 9.

$$x=\frac{-9±9\sqrt{37}}{18}$$

Now solve the equation x=\frac{-9±9\sqrt{37}}{18} when ± is plus. Add -9 to 9\sqrt{37}.

$$x=\frac{9\sqrt{37}-9}{18}$$

Divide -9+9\sqrt{37} by 18.

$$x=\frac{\sqrt{37}-1}{2}$$

Now solve the equation x=\frac{-9±9\sqrt{37}}{18} when ± is minus. Subtract 9\sqrt{37} from -9.

$$x=\frac{-9\sqrt{37}-9}{18}$$

Divide -9-9\sqrt{37} by 18.

$$x=\frac{-\sqrt{37}-1}{2}$$

The equation is now solved.

$$x=\frac{\sqrt{37}-1}{2} x=\frac{-\sqrt{37}-1}{2}$$

Steps for Completing the Square

Use the distributive property to multiply 3x by 3x+1.

$$6x+9x^{2}+3x+9=90$$

Combine 6x and 3x to get 9x.

$$9x+9x^{2}+9=90$$

Subtract 9 from both sides.

$$9x+9x^{2}=90-9$$

Subtract 9 from 90 to get 81.

$$9x+9x^{2}=81$$

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.

$$9x^{2}+9x=81$$

Divide both sides by 9.

$$\frac{9x^{2}+9x}{9}=\frac{81}{9}$$

Dividing by 9 undoes the multiplication by 9.

$$x^{2}+\frac{9}{9}x=\frac{81}{9}$$

Divide 9 by 9.

$$x^{2}+x=\frac{81}{9}$$

Divide 81 by 9.

$$x^{2}+x=9$$

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

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

Square \frac{1}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+x+\frac{1}{4}=9+\frac{1}{4}$$

Add 9 to \frac{1}{4}.

$$x^{2}+x+\frac{1}{4}=\frac{37}{4}$$

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

$$\left(x+\frac{1}{2}\right)^{2}=\frac{37}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{1}{2}\right)^{2}}=\sqrt{\frac{37}{4}}$$

Simplify.

$$x+\frac{1}{2}=\frac{\sqrt{37}}{2} x+\frac{1}{2}=-\frac{\sqrt{37}}{2}$$

Subtract \frac{1}{2} from both sides of the equation.

$$x=\frac{\sqrt{37}-1}{2} x=\frac{-\sqrt{37}-1}{2}$$