Solve 16x^2+2x-3=0

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

Solve for x

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

$$x=\frac{3}{8}=0.375$$

Steps Using Factoring By Grouping

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 16x^{2}+ax+bx-3. To find a and b, set up a system to be solved.

$$a+b=2 ab=16\left(-3\right)=-48$$

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 -48.

$$-1,48 -2,24 -3,16 -4,12 -6,8$$

Calculate the sum for each pair.

$$-1+48=47 -2+24=22 -3+16=13 -4+12=8 -6+8=2$$

The solution is the pair that gives sum 2.

$$a=-6 b=8$$

Rewrite 16x^{2}+2x-3 as \left(16x^{2}-6x\right)+\left(8x-3\right).

$$\left(16x^{2}-6x\right)+\left(8x-3\right)$$

Factor out 2x in 16x^{2}-6x.

$$2x\left(8x-3\right)+8x-3$$

Factor out common term 8x-3 by using distributive property.

$$\left(8x-3\right)\left(2x+1\right)$$

To find equation solutions, solve 8x-3=0 and 2x+1=0.

$$x=\frac{3}{8} x=-\frac{1}{2}$$

Steps Using the Quadratic Formula

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.

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

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

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

Square 2.

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

Multiply -4 times 16.

$$x=\frac{-2±\sqrt{4-64\left(-3\right)}}{2\times 16}$$

Multiply -64 times -3.

$$x=\frac{-2±\sqrt{4+192}}{2\times 16}$$

Add 4 to 192.

$$x=\frac{-2±\sqrt{196}}{2\times 16}$$

Take the square root of 196.

$$x=\frac{-2±14}{2\times 16}$$

Multiply 2 times 16.

$$x=\frac{-2±14}{32}$$

Now solve the equation x=\frac{-2±14}{32} when ± is plus. Add -2 to 14.

$$x=\frac{12}{32}$$

Reduce the fraction \frac{12}{32} to lowest terms by extracting and canceling out 4.

$$x=\frac{3}{8}$$

Now solve the equation x=\frac{-2±14}{32} when ± is minus. Subtract 14 from -2.

$$x=-\frac{16}{32}$$

Reduce the fraction \frac{-16}{32} to lowest terms by extracting and canceling out 16.

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

The equation is now solved.

$$x=\frac{3}{8} x=-\frac{1}{2}$$

Steps for Completing the Square

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.

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

Add 3 to both sides of the equation.

$$16x^{2}+2x-3-\left(-3\right)=-\left(-3\right)$$

Subtracting -3 from itself leaves 0.

$$16x^{2}+2x=-\left(-3\right)$$

Subtract -3 from 0.

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

Divide both sides by 16.

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

Dividing by 16 undoes the multiplication by 16.

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

Reduce the fraction \frac{2}{16} to lowest terms by extracting and canceling out 2.

$$x^{2}+\frac{1}{8}x=\frac{3}{16}$$

Divide \frac{1}{8}, the coefficient of the x term, by 2 to get \frac{1}{16}. Then add the square of \frac{1}{16} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{1}{8}x+\left(\frac{1}{16}\right)^{2}=\frac{3}{16}+\left(\frac{1}{16}\right)^{2}$$

Square \frac{1}{16} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{1}{8}x+\frac{1}{256}=\frac{3}{16}+\frac{1}{256}$$

Add \frac{3}{16} to \frac{1}{256} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{1}{8}x+\frac{1}{256}=\frac{49}{256}$$

Factor x^{2}+\frac{1}{8}x+\frac{1}{256}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{1}{16}\right)^{2}=\frac{49}{256}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{1}{16}\right)^{2}}=\sqrt{\frac{49}{256}}$$

Simplify.

$$x+\frac{1}{16}=\frac{7}{16} x+\frac{1}{16}=-\frac{7}{16}$$

Subtract \frac{1}{16} from both sides of the equation.

$$x=\frac{3}{8} x=-\frac{1}{2}$$