Solve x=3(x+5)^2-4

$$x = 3 ( x + 5 ) ^ { 2 } - 4$$

Solve for x (complex solution)

$$x=\frac{-\sqrt{11}i-29}{6}\approx -4.833333333-0.552770798i$$

$$x=\frac{-29+\sqrt{11}i}{6}\approx -4.833333333+0.552770798i$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+5\right)^{2}.

$$x=3\left(x^{2}+10x+25\right)-4$$

Use the distributive property to multiply 3 by x^{2}+10x+25.

$$x=3x^{2}+30x+75-4$$

Subtract 4 from 75 to get 71.

$$x=3x^{2}+30x+71$$

Subtract 3x^{2} from both sides.

$$x-3x^{2}=30x+71$$

Subtract 30x from both sides.

$$x-3x^{2}-30x=71$$

Combine x and -30x to get -29x.

$$-29x-3x^{2}=71$$

Subtract 71 from both sides.

$$-29x-3x^{2}-71=0$$

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.

$$-3x^{2}-29x-71=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -3 for a, -29 for b, and -71 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-29\right)±\sqrt{\left(-29\right)^{2}-4\left(-3\right)\left(-71\right)}}{2\left(-3\right)}$$

Square -29.

$$x=\frac{-\left(-29\right)±\sqrt{841-4\left(-3\right)\left(-71\right)}}{2\left(-3\right)}$$

Multiply -4 times -3.

$$x=\frac{-\left(-29\right)±\sqrt{841+12\left(-71\right)}}{2\left(-3\right)}$$

Multiply 12 times -71.

$$x=\frac{-\left(-29\right)±\sqrt{841-852}}{2\left(-3\right)}$$

Add 841 to -852.

$$x=\frac{-\left(-29\right)±\sqrt{-11}}{2\left(-3\right)}$$

Take the square root of -11.

$$x=\frac{-\left(-29\right)±\sqrt{11}i}{2\left(-3\right)}$$

The opposite of -29 is 29.

$$x=\frac{29±\sqrt{11}i}{2\left(-3\right)}$$

Multiply 2 times -3.

$$x=\frac{29±\sqrt{11}i}{-6}$$

Now solve the equation x=\frac{29±\sqrt{11}i}{-6} when ± is plus. Add 29 to i\sqrt{11}.

$$x=\frac{29+\sqrt{11}i}{-6}$$

Divide 29+i\sqrt{11} by -6.

$$x=\frac{-\sqrt{11}i-29}{6}$$

Now solve the equation x=\frac{29±\sqrt{11}i}{-6} when ± is minus. Subtract i\sqrt{11} from 29.

$$x=\frac{-\sqrt{11}i+29}{-6}$$

Divide 29-i\sqrt{11} by -6.

$$x=\frac{-29+\sqrt{11}i}{6}$$

The equation is now solved.

$$x=\frac{-\sqrt{11}i-29}{6} x=\frac{-29+\sqrt{11}i}{6}$$

Steps for Completing the Square

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+5\right)^{2}.

$$x=3\left(x^{2}+10x+25\right)-4$$

Use the distributive property to multiply 3 by x^{2}+10x+25.

$$x=3x^{2}+30x+75-4$$

Subtract 4 from 75 to get 71.

$$x=3x^{2}+30x+71$$

Subtract 3x^{2} from both sides.

$$x-3x^{2}=30x+71$$

Subtract 30x from both sides.

$$x-3x^{2}-30x=71$$

Combine x and -30x to get -29x.

$$-29x-3x^{2}=71$$

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.

$$-3x^{2}-29x=71$$

Divide both sides by -3.

$$\frac{-3x^{2}-29x}{-3}=\frac{71}{-3}$$

Dividing by -3 undoes the multiplication by -3.

$$x^{2}+\left(-\frac{29}{-3}\right)x=\frac{71}{-3}$$

Divide -29 by -3.

$$x^{2}+\frac{29}{3}x=\frac{71}{-3}$$

Divide 71 by -3.

$$x^{2}+\frac{29}{3}x=-\frac{71}{3}$$

Divide \frac{29}{3}, the coefficient of the x term, by 2 to get \frac{29}{6}. Then add the square of \frac{29}{6} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{29}{3}x+\left(\frac{29}{6}\right)^{2}=-\frac{71}{3}+\left(\frac{29}{6}\right)^{2}$$

Square \frac{29}{6} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{29}{3}x+\frac{841}{36}=-\frac{71}{3}+\frac{841}{36}$$

Add -\frac{71}{3} to \frac{841}{36} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{29}{3}x+\frac{841}{36}=-\frac{11}{36}$$

Factor x^{2}+\frac{29}{3}x+\frac{841}{36}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{29}{6}\right)^{2}=-\frac{11}{36}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{29}{6}\right)^{2}}=\sqrt{-\frac{11}{36}}$$

Simplify.

$$x+\frac{29}{6}=\frac{\sqrt{11}i}{6} x+\frac{29}{6}=-\frac{\sqrt{11}i}{6}$$

Subtract \frac{29}{6} from both sides of the equation.

$$x=\frac{-29+\sqrt{11}i}{6} x=\frac{-\sqrt{11}i-29}{6}$$