Solve 8x^2=-19x-6

$$8 { x }^{ 2 } =-19x-6$$

Solve for x

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

$$x=-2$$

Steps Using Factoring By Grouping

Add 19x to both sides.

$$8x^{2}+19x=-6$$

Add 6 to both sides.

$$8x^{2}+19x+6=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 8x^{2}+ax+bx+6. To find a and b, set up a system to be solved.

$$a+b=19 ab=8\times 6=48$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. 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.

$$1+48=49 2+24=26 3+16=19 4+12=16 6+8=14$$

The solution is the pair that gives sum 19.

$$a=3 b=16$$

Rewrite 8x^{2}+19x+6 as \left(8x^{2}+3x\right)+\left(16x+6\right).

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

Factor out x in the first and 2 in the second group.

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

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

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

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

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

Steps Using the Quadratic Formula

Add 19x to both sides.

$$8x^{2}+19x=-6$$

Add 6 to both sides.

$$8x^{2}+19x+6=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 8 for a, 19 for b, and 6 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-19±\sqrt{19^{2}-4\times 8\times 6}}{2\times 8}$$

Square 19.

$$x=\frac{-19±\sqrt{361-4\times 8\times 6}}{2\times 8}$$

Multiply -4 times 8.

$$x=\frac{-19±\sqrt{361-32\times 6}}{2\times 8}$$

Multiply -32 times 6.

$$x=\frac{-19±\sqrt{361-192}}{2\times 8}$$

Add 361 to -192.

$$x=\frac{-19±\sqrt{169}}{2\times 8}$$

Take the square root of 169.

$$x=\frac{-19±13}{2\times 8}$$

Multiply 2 times 8.

$$x=\frac{-19±13}{16}$$

Now solve the equation x=\frac{-19±13}{16} when ± is plus. Add -19 to 13.

$$x=-\frac{6}{16}$$

Reduce the fraction \frac{-6}{16} to lowest terms by extracting and canceling out 2.

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

Now solve the equation x=\frac{-19±13}{16} when ± is minus. Subtract 13 from -19.

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

Divide -32 by 16.

$$x=-2$$

The equation is now solved.

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

Steps for Completing the Square

Add 19x to both sides.

$$8x^{2}+19x=-6$$

Divide both sides by 8.

$$\frac{8x^{2}+19x}{8}=-\frac{6}{8}$$

Dividing by 8 undoes the multiplication by 8.

$$x^{2}+\frac{19}{8}x=-\frac{6}{8}$$

Reduce the fraction \frac{-6}{8} to lowest terms by extracting and canceling out 2.

$$x^{2}+\frac{19}{8}x=-\frac{3}{4}$$

Divide \frac{19}{8}, the coefficient of the x term, by 2 to get \frac{19}{16}. Then add the square of \frac{19}{16} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{19}{8}x+\left(\frac{19}{16}\right)^{2}=-\frac{3}{4}+\left(\frac{19}{16}\right)^{2}$$

Square \frac{19}{16} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{19}{8}x+\frac{361}{256}=-\frac{3}{4}+\frac{361}{256}$$

Add -\frac{3}{4} to \frac{361}{256} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{19}{8}x+\frac{361}{256}=\frac{169}{256}$$

Factor x^{2}+\frac{19}{8}x+\frac{361}{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}.

$$\left(x+\frac{19}{16}\right)^{2}=\frac{169}{256}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{19}{16}\right)^{2}}=\sqrt{\frac{169}{256}}$$

Simplify.

$$x+\frac{19}{16}=\frac{13}{16} x+\frac{19}{16}=-\frac{13}{16}$$

Subtract \frac{19}{16} from both sides of the equation.

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