Solve 8x^2-48=-40x

$$8 x ^ { 2 } - 48 = - 40 x$$

Solve for x

$$x=-6$$

$$x=1$$

Steps Using Factoring By Grouping

Add 40x to both sides.

$$8x^{2}-48+40x=0$$

Divide both sides by 8.

$$x^{2}-6+5x=0$$

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

$$x^{2}+5x-6=0$$

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

$$a+b=5 ab=1\left(-6\right)=-6$$

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 -6.

$$-1,6 -2,3$$

Calculate the sum for each pair.

$$-1+6=5 -2+3=1$$

The solution is the pair that gives sum 5.

$$a=-1 b=6$$

Rewrite x^{2}+5x-6 as \left(x^{2}-x\right)+\left(6x-6\right).

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

Factor out x in the first and 6 in the second group.

$$x\left(x-1\right)+6\left(x-1\right)$$

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

$$\left(x-1\right)\left(x+6\right)$$

To find equation solutions, solve x-1=0 and x+6=0.

$$x=1 x=-6$$

Steps Using the Quadratic Formula

Add 40x to both sides.

$$8x^{2}-48+40x=0$$

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.

$$8x^{2}+40x-48=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 8 for a, 40 for b, and -48 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-40±\sqrt{40^{2}-4\times 8\left(-48\right)}}{2\times 8}$$

Square 40.

$$x=\frac{-40±\sqrt{1600-4\times 8\left(-48\right)}}{2\times 8}$$

Multiply -4 times 8.

$$x=\frac{-40±\sqrt{1600-32\left(-48\right)}}{2\times 8}$$

Multiply -32 times -48.

$$x=\frac{-40±\sqrt{1600+1536}}{2\times 8}$$

Add 1600 to 1536.

$$x=\frac{-40±\sqrt{3136}}{2\times 8}$$

Take the square root of 3136.

$$x=\frac{-40±56}{2\times 8}$$

Multiply 2 times 8.

$$x=\frac{-40±56}{16}$$

Now solve the equation x=\frac{-40±56}{16} when ± is plus. Add -40 to 56.

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

Divide 16 by 16.

$$x=1$$

Now solve the equation x=\frac{-40±56}{16} when ± is minus. Subtract 56 from -40.

$$x=-\frac{96}{16}$$

Divide -96 by 16.

$$x=-6$$

The equation is now solved.

$$x=1 x=-6$$

Steps for Completing the Square

Add 40x to both sides.

$$8x^{2}-48+40x=0$$

Add 48 to both sides. Anything plus zero gives itself.

$$8x^{2}+40x=48$$

Divide both sides by 8.

$$\frac{8x^{2}+40x}{8}=\frac{48}{8}$$

Dividing by 8 undoes the multiplication by 8.

$$x^{2}+\frac{40}{8}x=\frac{48}{8}$$

Divide 40 by 8.

$$x^{2}+5x=\frac{48}{8}$$

Divide 48 by 8.

$$x^{2}+5x=6$$

Divide 5, the coefficient of the x term, by 2 to get \frac{5}{2}. Then add the square of \frac{5}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+5x+\left(\frac{5}{2}\right)^{2}=6+\left(\frac{5}{2}\right)^{2}$$

Square \frac{5}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+5x+\frac{25}{4}=6+\frac{25}{4}$$

Add 6 to \frac{25}{4}.

$$x^{2}+5x+\frac{25}{4}=\frac{49}{4}$$

Factor x^{2}+5x+\frac{25}{4}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{5}{2}\right)^{2}=\frac{49}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{5}{2}\right)^{2}}=\sqrt{\frac{49}{4}}$$

Simplify.

$$x+\frac{5}{2}=\frac{7}{2} x+\frac{5}{2}=-\frac{7}{2}$$

Subtract \frac{5}{2} from both sides of the equation.

$$x=1 x=-6$$