Solve 32x^2-17=-18x

$$32 x ^ { 2 } - 17 = - 18 x$$

Solve for x

$$x = -\frac{17}{16} = -1\frac{1}{16} = -1.0625$$

$$x=\frac{1}{2}=0.5$$

Steps Using Factoring By Grouping

Add 18x to both sides.

$$32x^{2}-17+18x=0$$

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$32x^{2}+18x-17=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 32x^{2}+ax+bx-17. To find a and b, set up a system to be solved.

$$a+b=18 ab=32\left(-17\right)=-544$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -544.

$$-1,544 -2,272 -4,136 -8,68 -16,34 -17,32$$

Calculate the sum for each pair.

$$-1+544=543 -2+272=270 -4+136=132 -8+68=60 -16+34=18 -17+32=15$$

The solution is the pair that gives sum 18.

$$a=-16 b=34$$

Rewrite 32x^{2}+18x-17 as \left(32x^{2}-16x\right)+\left(34x-17\right).

$$\left(32x^{2}-16x\right)+\left(34x-17\right)$$

Factor out 16x in the first and 17 in the second group.

$$16x\left(2x-1\right)+17\left(2x-1\right)$$

Factor out common term 2x-1 by using distributive property.

$$\left(2x-1\right)\left(16x+17\right)$$

To find equation solutions, solve 2x-1=0 and 16x+17=0.

$$x=\frac{1}{2} x=-\frac{17}{16}$$

Steps Using the Quadratic Formula

Add 18x to both sides.

$$32x^{2}-17+18x=0$$

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.

$$32x^{2}+18x-17=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 32 for a, 18 for b, and -17 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-18±\sqrt{18^{2}-4\times 32\left(-17\right)}}{2\times 32}$$

Square 18.

$$x=\frac{-18±\sqrt{324-4\times 32\left(-17\right)}}{2\times 32}$$

Multiply -4 times 32.

$$x=\frac{-18±\sqrt{324-128\left(-17\right)}}{2\times 32}$$

Multiply -128 times -17.

$$x=\frac{-18±\sqrt{324+2176}}{2\times 32}$$

Add 324 to 2176.

$$x=\frac{-18±\sqrt{2500}}{2\times 32}$$

Take the square root of 2500.

$$x=\frac{-18±50}{2\times 32}$$

Multiply 2 times 32.

$$x=\frac{-18±50}{64}$$

Now solve the equation x=\frac{-18±50}{64} when ± is plus. Add -18 to 50.

$$x=\frac{32}{64}$$

Reduce the fraction \frac{32}{64} to lowest terms by extracting and canceling out 32.

$$x=\frac{1}{2}$$

Now solve the equation x=\frac{-18±50}{64} when ± is minus. Subtract 50 from -18.

$$x=-\frac{68}{64}$$

Reduce the fraction \frac{-68}{64} to lowest terms by extracting and canceling out 4.

$$x=-\frac{17}{16}$$

The equation is now solved.

$$x=\frac{1}{2} x=-\frac{17}{16}$$

Steps for Completing the Square

Add 18x to both sides.

$$32x^{2}-17+18x=0$$

Add 17 to both sides. Anything plus zero gives itself.

$$32x^{2}+18x=17$$

Divide both sides by 32.

$$\frac{32x^{2}+18x}{32}=\frac{17}{32}$$

Dividing by 32 undoes the multiplication by 32.

$$x^{2}+\frac{18}{32}x=\frac{17}{32}$$

Reduce the fraction \frac{18}{32} to lowest terms by extracting and canceling out 2.

$$x^{2}+\frac{9}{16}x=\frac{17}{32}$$

Divide \frac{9}{16}, the coefficient of the x term, by 2 to get \frac{9}{32}. Then add the square of \frac{9}{32} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{9}{16}x+\left(\frac{9}{32}\right)^{2}=\frac{17}{32}+\left(\frac{9}{32}\right)^{2}$$

Square \frac{9}{32} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{9}{16}x+\frac{81}{1024}=\frac{17}{32}+\frac{81}{1024}$$

Add \frac{17}{32} to \frac{81}{1024} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{9}{16}x+\frac{81}{1024}=\frac{625}{1024}$$

Factor x^{2}+\frac{9}{16}x+\frac{81}{1024}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{9}{32}\right)^{2}=\frac{625}{1024}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{9}{32}\right)^{2}}=\sqrt{\frac{625}{1024}}$$

Simplify.

$$x+\frac{9}{32}=\frac{25}{32} x+\frac{9}{32}=-\frac{25}{32}$$

Subtract \frac{9}{32} from both sides of the equation.

$$x=\frac{1}{2} x=-\frac{17}{16}$$