Solve 36*4x+5=(3x+15)(3x+15)

$$36 \cdot 4 x + 5 = ( 3 x + 15 ) ( 3 x + 15 )$$

Solve for x (complex solution)

$$x=-\frac{\sqrt{139}i}{3}+3\approx 3-3.929942041i$$

$$x=\frac{\sqrt{139}i}{3}+3\approx 3+3.929942041i$$

Steps Using the Quadratic Formula

Multiply 3x+15 and 3x+15 to get \left(3x+15\right)^{2}.

$$36\times 4x+5=\left(3x+15\right)^{2}$$

Multiply 36 and 4 to get 144.

$$144x+5=\left(3x+15\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(3x+15\right)^{2}.

$$144x+5=9x^{2}+90x+225$$

Subtract 9x^{2} from both sides.

$$144x+5-9x^{2}=90x+225$$

Subtract 90x from both sides.

$$144x+5-9x^{2}-90x=225$$

Combine 144x and -90x to get 54x.

$$54x+5-9x^{2}=225$$

Subtract 225 from both sides.

$$54x+5-9x^{2}-225=0$$

Subtract 225 from 5 to get -220.

$$54x-220-9x^{2}=0$$

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.

$$-9x^{2}+54x-220=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -9 for a, 54 for b, and -220 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-54±\sqrt{54^{2}-4\left(-9\right)\left(-220\right)}}{2\left(-9\right)}$$

Square 54.

$$x=\frac{-54±\sqrt{2916-4\left(-9\right)\left(-220\right)}}{2\left(-9\right)}$$

Multiply -4 times -9.

$$x=\frac{-54±\sqrt{2916+36\left(-220\right)}}{2\left(-9\right)}$$

Multiply 36 times -220.

$$x=\frac{-54±\sqrt{2916-7920}}{2\left(-9\right)}$$

Add 2916 to -7920.

$$x=\frac{-54±\sqrt{-5004}}{2\left(-9\right)}$$

Take the square root of -5004.

$$x=\frac{-54±6\sqrt{139}i}{2\left(-9\right)}$$

Multiply 2 times -9.

$$x=\frac{-54±6\sqrt{139}i}{-18}$$

Now solve the equation x=\frac{-54±6\sqrt{139}i}{-18} when ± is plus. Add -54 to 6i\sqrt{139}.

$$x=\frac{-54+6\sqrt{139}i}{-18}$$

Divide -54+6i\sqrt{139} by -18.

$$x=-\frac{\sqrt{139}i}{3}+3$$

Now solve the equation x=\frac{-54±6\sqrt{139}i}{-18} when ± is minus. Subtract 6i\sqrt{139} from -54.

$$x=\frac{-6\sqrt{139}i-54}{-18}$$

Divide -54-6i\sqrt{139} by -18.

$$x=\frac{\sqrt{139}i}{3}+3$$

The equation is now solved.

$$x=-\frac{\sqrt{139}i}{3}+3 x=\frac{\sqrt{139}i}{3}+3$$

Steps for Completing the Square

Multiply 3x+15 and 3x+15 to get \left(3x+15\right)^{2}.

$$36\times 4x+5=\left(3x+15\right)^{2}$$

Multiply 36 and 4 to get 144.

$$144x+5=\left(3x+15\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(3x+15\right)^{2}.

$$144x+5=9x^{2}+90x+225$$

Subtract 9x^{2} from both sides.

$$144x+5-9x^{2}=90x+225$$

Subtract 90x from both sides.

$$144x+5-9x^{2}-90x=225$$

Combine 144x and -90x to get 54x.

$$54x+5-9x^{2}=225$$

Subtract 5 from both sides.

$$54x-9x^{2}=225-5$$

Subtract 5 from 225 to get 220.

$$54x-9x^{2}=220$$

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.

$$-9x^{2}+54x=220$$

Divide both sides by -9.

$$\frac{-9x^{2}+54x}{-9}=\frac{220}{-9}$$

Dividing by -9 undoes the multiplication by -9.

$$x^{2}+\frac{54}{-9}x=\frac{220}{-9}$$

Divide 54 by -9.

$$x^{2}-6x=\frac{220}{-9}$$

Divide 220 by -9.

$$x^{2}-6x=-\frac{220}{9}$$

Divide -6, the coefficient of the x term, by 2 to get -3. Then add the square of -3 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-6x+\left(-3\right)^{2}=-\frac{220}{9}+\left(-3\right)^{2}$$

Square -3.

$$x^{2}-6x+9=-\frac{220}{9}+9$$

Add -\frac{220}{9} to 9.

$$x^{2}-6x+9=-\frac{139}{9}$$

Factor x^{2}-6x+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}.

$$\left(x-3\right)^{2}=-\frac{139}{9}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-3\right)^{2}}=\sqrt{-\frac{139}{9}}$$

Simplify.

$$x-3=\frac{\sqrt{139}i}{3} x-3=-\frac{\sqrt{139}i}{3}$$

Add 3 to both sides of the equation.

$$x=\frac{\sqrt{139}i}{3}+3 x=-\frac{\sqrt{139}i}{3}+3$$