Solve 28x+8=-24x^2

$$28 x + 8 = - 24 x ^ { 2 }$$

Solve for x

$$x=-\frac{2}{3}\approx -0.666666667$$

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

Steps Using Factoring By Grouping

Add 24x^{2} to both sides.

$$28x+8+24x^{2}=0$$

Divide both sides by 4.

$$7x+2+6x^{2}=0$$

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

$$6x^{2}+7x+2=0$$

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

$$a+b=7 ab=6\times 2=12$$

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 12.

$$1,12 2,6 3,4$$

Calculate the sum for each pair.

$$1+12=13 2+6=8 3+4=7$$

The solution is the pair that gives sum 7.

$$a=3 b=4$$

Rewrite 6x^{2}+7x+2 as \left(6x^{2}+3x\right)+\left(4x+2\right).

$$\left(6x^{2}+3x\right)+\left(4x+2\right)$$

Factor out 3x in the first and 2 in the second group.

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

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

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

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

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

Steps Using the Quadratic Formula

Add 24x^{2} to both sides.

$$28x+8+24x^{2}=0$$

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.

$$24x^{2}+28x+8=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 24 for a, 28 for b, and 8 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-28±\sqrt{28^{2}-4\times 24\times 8}}{2\times 24}$$

Square 28.

$$x=\frac{-28±\sqrt{784-4\times 24\times 8}}{2\times 24}$$

Multiply -4 times 24.

$$x=\frac{-28±\sqrt{784-96\times 8}}{2\times 24}$$

Multiply -96 times 8.

$$x=\frac{-28±\sqrt{784-768}}{2\times 24}$$

Add 784 to -768.

$$x=\frac{-28±\sqrt{16}}{2\times 24}$$

Take the square root of 16.

$$x=\frac{-28±4}{2\times 24}$$

Multiply 2 times 24.

$$x=\frac{-28±4}{48}$$

Now solve the equation x=\frac{-28±4}{48} when ± is plus. Add -28 to 4.

$$x=-\frac{24}{48}$$

Reduce the fraction \frac{-24}{48} to lowest terms by extracting and canceling out 24.

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

Now solve the equation x=\frac{-28±4}{48} when ± is minus. Subtract 4 from -28.

$$x=-\frac{32}{48}$$

Reduce the fraction \frac{-32}{48} to lowest terms by extracting and canceling out 16.

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

The equation is now solved.

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

Steps for Completing the Square

Add 24x^{2} to both sides.

$$28x+8+24x^{2}=0$$

Subtract 8 from both sides. Anything subtracted from zero gives its negation.

$$28x+24x^{2}=-8$$

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.

$$24x^{2}+28x=-8$$

Divide both sides by 24.

$$\frac{24x^{2}+28x}{24}=-\frac{8}{24}$$

Dividing by 24 undoes the multiplication by 24.

$$x^{2}+\frac{28}{24}x=-\frac{8}{24}$$

Reduce the fraction \frac{28}{24} to lowest terms by extracting and canceling out 4.

$$x^{2}+\frac{7}{6}x=-\frac{8}{24}$$

Reduce the fraction \frac{-8}{24} to lowest terms by extracting and canceling out 8.

$$x^{2}+\frac{7}{6}x=-\frac{1}{3}$$

Divide \frac{7}{6}, the coefficient of the x term, by 2 to get \frac{7}{12}. Then add the square of \frac{7}{12} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{7}{6}x+\left(\frac{7}{12}\right)^{2}=-\frac{1}{3}+\left(\frac{7}{12}\right)^{2}$$

Square \frac{7}{12} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{7}{6}x+\frac{49}{144}=-\frac{1}{3}+\frac{49}{144}$$

Add -\frac{1}{3} to \frac{49}{144} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{7}{6}x+\frac{49}{144}=\frac{1}{144}$$

Factor x^{2}+\frac{7}{6}x+\frac{49}{144}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{7}{12}\right)^{2}=\frac{1}{144}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{7}{12}\right)^{2}}=\sqrt{\frac{1}{144}}$$

Simplify.

$$x+\frac{7}{12}=\frac{1}{12} x+\frac{7}{12}=-\frac{1}{12}$$

Subtract \frac{7}{12} from both sides of the equation.

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