Solve 6x^2+3x+(3x+3^2)=90

$$6 { x }^{ 2 } +3x+(3x+ { 3 }^{ 2 } )=90$$

Solve for x

$$x = \frac{\sqrt{55} - 1}{2} \approx 3.208099244$$

$$x=\frac{-\sqrt{55}-1}{2}\approx -4.208099244$$

Steps Using the Quadratic Formula

Combine 3x and 3x to get 6x.

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

Calculate 3 to the power of 2 and get 9.

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

Subtract 90 from both sides.

$$6x^{2}+6x+9-90=0$$

Subtract 90 from 9 to get -81.

$$6x^{2}+6x-81=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 6 for a, 6 for b, and -81 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-6±\sqrt{6^{2}-4\times 6\left(-81\right)}}{2\times 6}$$

Square 6.

$$x=\frac{-6±\sqrt{36-4\times 6\left(-81\right)}}{2\times 6}$$

Multiply -4 times 6.

$$x=\frac{-6±\sqrt{36-24\left(-81\right)}}{2\times 6}$$

Multiply -24 times -81.

$$x=\frac{-6±\sqrt{36+1944}}{2\times 6}$$

Add 36 to 1944.

$$x=\frac{-6±\sqrt{1980}}{2\times 6}$$

Take the square root of 1980.

$$x=\frac{-6±6\sqrt{55}}{2\times 6}$$

Multiply 2 times 6.

$$x=\frac{-6±6\sqrt{55}}{12}$$

Now solve the equation x=\frac{-6±6\sqrt{55}}{12} when ± is plus. Add -6 to 6\sqrt{55}.

$$x=\frac{6\sqrt{55}-6}{12}$$

Divide -6+6\sqrt{55} by 12.

$$x=\frac{\sqrt{55}-1}{2}$$

Now solve the equation x=\frac{-6±6\sqrt{55}}{12} when ± is minus. Subtract 6\sqrt{55} from -6.

$$x=\frac{-6\sqrt{55}-6}{12}$$

Divide -6-6\sqrt{55} by 12.

$$x=\frac{-\sqrt{55}-1}{2}$$

The equation is now solved.

$$x=\frac{\sqrt{55}-1}{2} x=\frac{-\sqrt{55}-1}{2}$$

Steps for Completing the Square

Combine 3x and 3x to get 6x.

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

Calculate 3 to the power of 2 and get 9.

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

Subtract 9 from both sides.

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

Subtract 9 from 90 to get 81.

$$6x^{2}+6x=81$$

Divide both sides by 6.

$$\frac{6x^{2}+6x}{6}=\frac{81}{6}$$

Dividing by 6 undoes the multiplication by 6.

$$x^{2}+\frac{6}{6}x=\frac{81}{6}$$

Divide 6 by 6.

$$x^{2}+x=\frac{81}{6}$$

Reduce the fraction \frac{81}{6} to lowest terms by extracting and canceling out 3.

$$x^{2}+x=\frac{27}{2}$$

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.

$$x^{2}+x+\left(\frac{1}{2}\right)^{2}=\frac{27}{2}+\left(\frac{1}{2}\right)^{2}$$

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

$$x^{2}+x+\frac{1}{4}=\frac{27}{2}+\frac{1}{4}$$

Add \frac{27}{2} to \frac{1}{4} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+x+\frac{1}{4}=\frac{55}{4}$$

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}.

$$\left(x+\frac{1}{2}\right)^{2}=\frac{55}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{1}{2}\right)^{2}}=\sqrt{\frac{55}{4}}$$

Simplify.

$$x+\frac{1}{2}=\frac{\sqrt{55}}{2} x+\frac{1}{2}=-\frac{\sqrt{55}}{2}$$

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

$$x=\frac{\sqrt{55}-1}{2} x=\frac{-\sqrt{55}-1}{2}$$