Solve 15=3+36x-16x^2

$$15=3+36x-16 { x }^{ 2 }$$

Solve for x

$$x = \frac{\sqrt{33} + 9}{8} \approx 1.843070331$$

$$x=\frac{9-\sqrt{33}}{8}\approx 0.406929669$$

Steps Using the Quadratic Formula

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

$$3+36x-16x^{2}=15$$

Subtract 15 from both sides.

$$3+36x-16x^{2}-15=0$$

Subtract 15 from 3 to get -12.

$$-12+36x-16x^{2}=0$$

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.

$$-16x^{2}+36x-12=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -16 for a, 36 for b, and -12 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-36±\sqrt{36^{2}-4\left(-16\right)\left(-12\right)}}{2\left(-16\right)}$$

Square 36.

$$x=\frac{-36±\sqrt{1296-4\left(-16\right)\left(-12\right)}}{2\left(-16\right)}$$

Multiply -4 times -16.

$$x=\frac{-36±\sqrt{1296+64\left(-12\right)}}{2\left(-16\right)}$$

Multiply 64 times -12.

$$x=\frac{-36±\sqrt{1296-768}}{2\left(-16\right)}$$

Add 1296 to -768.

$$x=\frac{-36±\sqrt{528}}{2\left(-16\right)}$$

Take the square root of 528.

$$x=\frac{-36±4\sqrt{33}}{2\left(-16\right)}$$

Multiply 2 times -16.

$$x=\frac{-36±4\sqrt{33}}{-32}$$

Now solve the equation x=\frac{-36±4\sqrt{33}}{-32} when ± is plus. Add -36 to 4\sqrt{33}.

$$x=\frac{4\sqrt{33}-36}{-32}$$

Divide -36+4\sqrt{33} by -32.

$$x=\frac{9-\sqrt{33}}{8}$$

Now solve the equation x=\frac{-36±4\sqrt{33}}{-32} when ± is minus. Subtract 4\sqrt{33} from -36.

$$x=\frac{-4\sqrt{33}-36}{-32}$$

Divide -36-4\sqrt{33} by -32.

$$x=\frac{\sqrt{33}+9}{8}$$

The equation is now solved.

$$x=\frac{9-\sqrt{33}}{8} x=\frac{\sqrt{33}+9}{8}$$

Steps for Completing the Square

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

$$3+36x-16x^{2}=15$$

Subtract 3 from both sides.

$$36x-16x^{2}=15-3$$

Subtract 3 from 15 to get 12.

$$36x-16x^{2}=12$$

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.

$$-16x^{2}+36x=12$$

Divide both sides by -16.

$$\frac{-16x^{2}+36x}{-16}=\frac{12}{-16}$$

Dividing by -16 undoes the multiplication by -16.

$$x^{2}+\frac{36}{-16}x=\frac{12}{-16}$$

Reduce the fraction \frac{36}{-16} to lowest terms by extracting and canceling out 4.

$$x^{2}-\frac{9}{4}x=\frac{12}{-16}$$

Reduce the fraction \frac{12}{-16} to lowest terms by extracting and canceling out 4.

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

Divide -\frac{9}{4}, the coefficient of the x term, by 2 to get -\frac{9}{8}. Then add the square of -\frac{9}{8} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{9}{4}x+\left(-\frac{9}{8}\right)^{2}=-\frac{3}{4}+\left(-\frac{9}{8}\right)^{2}$$

Square -\frac{9}{8} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{9}{4}x+\frac{81}{64}=-\frac{3}{4}+\frac{81}{64}$$

Add -\frac{3}{4} to \frac{81}{64} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{9}{4}x+\frac{81}{64}=\frac{33}{64}$$

Factor x^{2}-\frac{9}{4}x+\frac{81}{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}.

$$\left(x-\frac{9}{8}\right)^{2}=\frac{33}{64}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{9}{8}\right)^{2}}=\sqrt{\frac{33}{64}}$$

Simplify.

$$x-\frac{9}{8}=\frac{\sqrt{33}}{8} x-\frac{9}{8}=-\frac{\sqrt{33}}{8}$$

Add \frac{9}{8} to both sides of the equation.

$$x=\frac{\sqrt{33}+9}{8} x=\frac{9-\sqrt{33}}{8}$$