Solve 16.9=12x-4x^2

$$16.9 = 12 x - 4 x ^ { 2 }$$

Solve for x (complex solution)

$$x=\frac{i\sqrt{790}}{20}+1.5\approx 1.5+1.405346932i$$

$$x=-\frac{i\sqrt{790}}{20}+1.5\approx 1.5-1.405346932i$$

Steps Using the Quadratic Formula

Swap sides so that all variable terms are on the left hand side.

$$12x-4x^{2}=16.9$$

Subtract 16.9 from both sides.

$$12x-4x^{2}-16.9=0$$

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.

$$-4x^{2}+12x-16.9=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -4 for a, 12 for b, and -16.9 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-12±\sqrt{12^{2}-4\left(-4\right)\left(-16.9\right)}}{2\left(-4\right)}$$

Square 12.

$$x=\frac{-12±\sqrt{144-4\left(-4\right)\left(-16.9\right)}}{2\left(-4\right)}$$

Multiply -4 times -4.

$$x=\frac{-12±\sqrt{144+16\left(-16.9\right)}}{2\left(-4\right)}$$

Multiply 16 times -16.9.

$$x=\frac{-12±\sqrt{144-270.4}}{2\left(-4\right)}$$

Add 144 to -270.4.

$$x=\frac{-12±\sqrt{-126.4}}{2\left(-4\right)}$$

Take the square root of -126.4.

$$x=\frac{-12±\frac{2\sqrt{790}i}{5}}{2\left(-4\right)}$$

Multiply 2 times -4.

$$x=\frac{-12±\frac{2\sqrt{790}i}{5}}{-8}$$

Now solve the equation x=\frac{-12±\frac{2\sqrt{790}i}{5}}{-8} when ± is plus. Add -12 to \frac{2i\sqrt{790}}{5}.

$$x=\frac{\frac{2\sqrt{790}i}{5}-12}{-8}$$

Divide -12+\frac{2i\sqrt{790}}{5} by -8.

$$x=-\frac{\sqrt{790}i}{20}+\frac{3}{2}$$

Now solve the equation x=\frac{-12±\frac{2\sqrt{790}i}{5}}{-8} when ± is minus. Subtract \frac{2i\sqrt{790}}{5} from -12.

$$x=\frac{-\frac{2\sqrt{790}i}{5}-12}{-8}$$

Divide -12-\frac{2i\sqrt{790}}{5} by -8.

$$x=\frac{\sqrt{790}i}{20}+\frac{3}{2}$$

The equation is now solved.

$$x=-\frac{\sqrt{790}i}{20}+\frac{3}{2} x=\frac{\sqrt{790}i}{20}+\frac{3}{2}$$

Steps for Completing the Square

Swap sides so that all variable terms are on the left hand side.

$$12x-4x^{2}=16.9$$

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.

$$-4x^{2}+12x=16.9$$

Divide both sides by -4.

$$\frac{-4x^{2}+12x}{-4}=\frac{16.9}{-4}$$

Dividing by -4 undoes the multiplication by -4.

$$x^{2}+\frac{12}{-4}x=\frac{16.9}{-4}$$

Divide 12 by -4.

$$x^{2}-3x=\frac{16.9}{-4}$$

Divide 16.9 by -4.

$$x^{2}-3x=-4.225$$

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

$$x^{2}-3x+\left(-\frac{3}{2}\right)^{2}=-4.225+\left(-\frac{3}{2}\right)^{2}$$

Square -\frac{3}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-3x+\frac{9}{4}=-4.225+\frac{9}{4}$$

Add -4.225 to \frac{9}{4} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-3x+\frac{9}{4}=-\frac{79}{40}$$

Factor x^{2}-3x+\frac{9}{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}.

$$\left(x-\frac{3}{2}\right)^{2}=-\frac{79}{40}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{3}{2}\right)^{2}}=\sqrt{-\frac{79}{40}}$$

Simplify.

$$x-\frac{3}{2}=\frac{\sqrt{790}i}{20} x-\frac{3}{2}=-\frac{\sqrt{790}i}{20}$$

Add \frac{3}{2} to both sides of the equation.

$$x=\frac{\sqrt{790}i}{20}+\frac{3}{2} x=-\frac{\sqrt{790}i}{20}+\frac{3}{2}$$