Solve 3x=4x^2+16

$$3 x = 4 x ^ { 2 } + 16$$

Solve for x (complex solution)

$$x=\frac{-\sqrt{247}i+3}{8}\approx 0.375-1.964529206i$$

$$x=\frac{3+\sqrt{247}i}{8}\approx 0.375+1.964529206i$$

Steps Using the Quadratic Formula

Subtract 4x^{2} from both sides.

$$3x-4x^{2}=16$$

Subtract 16 from both sides.

$$3x-4x^{2}-16=0$$

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.

$$-4x^{2}+3x-16=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -4 for a, 3 for b, and -16 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-3±\sqrt{3^{2}-4\left(-4\right)\left(-16\right)}}{2\left(-4\right)}$$

Square 3.

$$x=\frac{-3±\sqrt{9-4\left(-4\right)\left(-16\right)}}{2\left(-4\right)}$$

Multiply -4 times -4.

$$x=\frac{-3±\sqrt{9+16\left(-16\right)}}{2\left(-4\right)}$$

Multiply 16 times -16.

$$x=\frac{-3±\sqrt{9-256}}{2\left(-4\right)}$$

Add 9 to -256.

$$x=\frac{-3±\sqrt{-247}}{2\left(-4\right)}$$

Take the square root of -247.

$$x=\frac{-3±\sqrt{247}i}{2\left(-4\right)}$$

Multiply 2 times -4.

$$x=\frac{-3±\sqrt{247}i}{-8}$$

Now solve the equation x=\frac{-3±\sqrt{247}i}{-8} when ± is plus. Add -3 to i\sqrt{247}.

$$x=\frac{-3+\sqrt{247}i}{-8}$$

Divide -3+i\sqrt{247} by -8.

$$x=\frac{-\sqrt{247}i+3}{8}$$

Now solve the equation x=\frac{-3±\sqrt{247}i}{-8} when ± is minus. Subtract i\sqrt{247} from -3.

$$x=\frac{-\sqrt{247}i-3}{-8}$$

Divide -3-i\sqrt{247} by -8.

$$x=\frac{3+\sqrt{247}i}{8}$$

The equation is now solved.

$$x=\frac{-\sqrt{247}i+3}{8} x=\frac{3+\sqrt{247}i}{8}$$

Steps for Completing the Square

Subtract 4x^{2} from both sides.

$$3x-4x^{2}=16$$

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.

$$-4x^{2}+3x=16$$

Divide both sides by -4.

$$\frac{-4x^{2}+3x}{-4}=\frac{16}{-4}$$

Dividing by -4 undoes the multiplication by -4.

$$x^{2}+\frac{3}{-4}x=\frac{16}{-4}$$

Divide 3 by -4.

$$x^{2}-\frac{3}{4}x=\frac{16}{-4}$$

Divide 16 by -4.

$$x^{2}-\frac{3}{4}x=-4$$

Divide -\frac{3}{4}, the coefficient of the x term, by 2 to get -\frac{3}{8}. Then add the square of -\frac{3}{8} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{3}{4}x+\left(-\frac{3}{8}\right)^{2}=-4+\left(-\frac{3}{8}\right)^{2}$$

Square -\frac{3}{8} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{3}{4}x+\frac{9}{64}=-4+\frac{9}{64}$$

Add -4 to \frac{9}{64}.

$$x^{2}-\frac{3}{4}x+\frac{9}{64}=-\frac{247}{64}$$

Factor x^{2}-\frac{3}{4}x+\frac{9}{64}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{3}{8}\right)^{2}=-\frac{247}{64}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{3}{8}\right)^{2}}=\sqrt{-\frac{247}{64}}$$

Simplify.

$$x-\frac{3}{8}=\frac{\sqrt{247}i}{8} x-\frac{3}{8}=-\frac{\sqrt{247}i}{8}$$

Add \frac{3}{8} to both sides of the equation.

$$x=\frac{3+\sqrt{247}i}{8} x=\frac{-\sqrt{247}i+3}{8}$$