Solve x=-3x^2+16x+12

$$x = - 3 x ^ { 2 } + 16 x + 12$$

Solve for x

$$x = \frac{\sqrt{41} + 5}{2} \approx 5.701562119$$

$$x=\frac{5-\sqrt{41}}{2}\approx -0.701562119$$

Steps Using the Quadratic Formula

Add 3x^{2} to both sides.

$$x+3x^{2}=16x+12$$

Subtract 16x from both sides.

$$x+3x^{2}-16x=12$$

Combine x and -16x to get -15x.

$$-15x+3x^{2}=12$$

Subtract 12 from both sides.

$$-15x+3x^{2}-12=0$$

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.

$$3x^{2}-15x-12=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, -15 for b, and -12 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-15\right)±\sqrt{\left(-15\right)^{2}-4\times 3\left(-12\right)}}{2\times 3}$$

Square -15.

$$x=\frac{-\left(-15\right)±\sqrt{225-4\times 3\left(-12\right)}}{2\times 3}$$

Multiply -4 times 3.

$$x=\frac{-\left(-15\right)±\sqrt{225-12\left(-12\right)}}{2\times 3}$$

Multiply -12 times -12.

$$x=\frac{-\left(-15\right)±\sqrt{225+144}}{2\times 3}$$

Add 225 to 144.

$$x=\frac{-\left(-15\right)±\sqrt{369}}{2\times 3}$$

Take the square root of 369.

$$x=\frac{-\left(-15\right)±3\sqrt{41}}{2\times 3}$$

The opposite of -15 is 15.

$$x=\frac{15±3\sqrt{41}}{2\times 3}$$

Multiply 2 times 3.

$$x=\frac{15±3\sqrt{41}}{6}$$

Now solve the equation x=\frac{15±3\sqrt{41}}{6} when ± is plus. Add 15 to 3\sqrt{41}.

$$x=\frac{3\sqrt{41}+15}{6}$$

Divide 15+3\sqrt{41} by 6.

$$x=\frac{\sqrt{41}+5}{2}$$

Now solve the equation x=\frac{15±3\sqrt{41}}{6} when ± is minus. Subtract 3\sqrt{41} from 15.

$$x=\frac{15-3\sqrt{41}}{6}$$

Divide 15-3\sqrt{41} by 6.

$$x=\frac{5-\sqrt{41}}{2}$$

The equation is now solved.

$$x=\frac{\sqrt{41}+5}{2} x=\frac{5-\sqrt{41}}{2}$$

Steps for Completing the Square

Add 3x^{2} to both sides.

$$x+3x^{2}=16x+12$$

Subtract 16x from both sides.

$$x+3x^{2}-16x=12$$

Combine x and -16x to get -15x.

$$-15x+3x^{2}=12$$

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.

$$3x^{2}-15x=12$$

Divide both sides by 3.

$$\frac{3x^{2}-15x}{3}=\frac{12}{3}$$

Dividing by 3 undoes the multiplication by 3.

$$x^{2}+\left(-\frac{15}{3}\right)x=\frac{12}{3}$$

Divide -15 by 3.

$$x^{2}-5x=\frac{12}{3}$$

Divide 12 by 3.

$$x^{2}-5x=4$$

Divide -5, the coefficient of the x term, by 2 to get -\frac{5}{2}. Then add the square of -\frac{5}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-5x+\left(-\frac{5}{2}\right)^{2}=4+\left(-\frac{5}{2}\right)^{2}$$

Square -\frac{5}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-5x+\frac{25}{4}=4+\frac{25}{4}$$

Add 4 to \frac{25}{4}.

$$x^{2}-5x+\frac{25}{4}=\frac{41}{4}$$

Factor x^{2}-5x+\frac{25}{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}.

$$\left(x-\frac{5}{2}\right)^{2}=\frac{41}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{5}{2}\right)^{2}}=\sqrt{\frac{41}{4}}$$

Simplify.

$$x-\frac{5}{2}=\frac{\sqrt{41}}{2} x-\frac{5}{2}=-\frac{\sqrt{41}}{2}$$

Add \frac{5}{2} to both sides of the equation.

$$x=\frac{\sqrt{41}+5}{2} x=\frac{5-\sqrt{41}}{2}$$