Solve x^2+x-6=4x^2+3x+64

$$x ^ { 2 } + x - 6 = 4 x ^ { 2 } + 3 x + 64$$

Solve for x (complex solution)

$$x=\frac{-\sqrt{209}i-1}{3}\approx -0.333333333-4.818944098i$$

$$x=\frac{-1+\sqrt{209}i}{3}\approx -0.333333333+4.818944098i$$

Steps Using the Quadratic Formula

Subtract 4x^{2} from both sides.

$$x^{2}+x-6-4x^{2}=3x+64$$

Combine x^{2} and -4x^{2} to get -3x^{2}.

$$-3x^{2}+x-6=3x+64$$

Subtract 3x from both sides.

$$-3x^{2}+x-6-3x=64$$

Combine x and -3x to get -2x.

$$-3x^{2}-2x-6=64$$

Subtract 64 from both sides.

$$-3x^{2}-2x-6-64=0$$

Subtract 64 from -6 to get -70.

$$-3x^{2}-2x-70=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -3 for a, -2 for b, and -70 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-2\right)±\sqrt{\left(-2\right)^{2}-4\left(-3\right)\left(-70\right)}}{2\left(-3\right)}$$

Square -2.

$$x=\frac{-\left(-2\right)±\sqrt{4-4\left(-3\right)\left(-70\right)}}{2\left(-3\right)}$$

Multiply -4 times -3.

$$x=\frac{-\left(-2\right)±\sqrt{4+12\left(-70\right)}}{2\left(-3\right)}$$

Multiply 12 times -70.

$$x=\frac{-\left(-2\right)±\sqrt{4-840}}{2\left(-3\right)}$$

Add 4 to -840.

$$x=\frac{-\left(-2\right)±\sqrt{-836}}{2\left(-3\right)}$$

Take the square root of -836.

$$x=\frac{-\left(-2\right)±2\sqrt{209}i}{2\left(-3\right)}$$

The opposite of -2 is 2.

$$x=\frac{2±2\sqrt{209}i}{2\left(-3\right)}$$

Multiply 2 times -3.

$$x=\frac{2±2\sqrt{209}i}{-6}$$

Now solve the equation x=\frac{2±2\sqrt{209}i}{-6} when ± is plus. Add 2 to 2i\sqrt{209}.

$$x=\frac{2+2\sqrt{209}i}{-6}$$

Divide 2+2i\sqrt{209} by -6.

$$x=\frac{-\sqrt{209}i-1}{3}$$

Now solve the equation x=\frac{2±2\sqrt{209}i}{-6} when ± is minus. Subtract 2i\sqrt{209} from 2.

$$x=\frac{-2\sqrt{209}i+2}{-6}$$

Divide 2-2i\sqrt{209} by -6.

$$x=\frac{-1+\sqrt{209}i}{3}$$

The equation is now solved.

$$x=\frac{-\sqrt{209}i-1}{3} x=\frac{-1+\sqrt{209}i}{3}$$

Steps for Completing the Square

Subtract 4x^{2} from both sides.

$$x^{2}+x-6-4x^{2}=3x+64$$

Combine x^{2} and -4x^{2} to get -3x^{2}.

$$-3x^{2}+x-6=3x+64$$

Subtract 3x from both sides.

$$-3x^{2}+x-6-3x=64$$

Combine x and -3x to get -2x.

$$-3x^{2}-2x-6=64$$

Add 6 to both sides.

$$-3x^{2}-2x=64+6$$

Add 64 and 6 to get 70.

$$-3x^{2}-2x=70$$

Divide both sides by -3.

$$\frac{-3x^{2}-2x}{-3}=\frac{70}{-3}$$

Dividing by -3 undoes the multiplication by -3.

$$x^{2}+\left(-\frac{2}{-3}\right)x=\frac{70}{-3}$$

Divide -2 by -3.

$$x^{2}+\frac{2}{3}x=\frac{70}{-3}$$

Divide 70 by -3.

$$x^{2}+\frac{2}{3}x=-\frac{70}{3}$$

Divide \frac{2}{3}, the coefficient of the x term, by 2 to get \frac{1}{3}. Then add the square of \frac{1}{3} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{2}{3}x+\left(\frac{1}{3}\right)^{2}=-\frac{70}{3}+\left(\frac{1}{3}\right)^{2}$$

Square \frac{1}{3} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{2}{3}x+\frac{1}{9}=-\frac{70}{3}+\frac{1}{9}$$

Add -\frac{70}{3} to \frac{1}{9} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{2}{3}x+\frac{1}{9}=-\frac{209}{9}$$

Factor x^{2}+\frac{2}{3}x+\frac{1}{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}.

$$\left(x+\frac{1}{3}\right)^{2}=-\frac{209}{9}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{1}{3}\right)^{2}}=\sqrt{-\frac{209}{9}}$$

Simplify.

$$x+\frac{1}{3}=\frac{\sqrt{209}i}{3} x+\frac{1}{3}=-\frac{\sqrt{209}i}{3}$$

Subtract \frac{1}{3} from both sides of the equation.

$$x=\frac{-1+\sqrt{209}i}{3} x=\frac{-\sqrt{209}i-1}{3}$$