Solve x(x+1)*8=96

$$x ( x + 1 ) \times 8 = 96$$

Solve for x

$$x=3$$

$$x=-4$$

Steps Using the Quadratic Formula

Use the distributive property to multiply x by x+1.

$$\left(x^{2}+x\right)\times 8=96$$

Use the distributive property to multiply x^{2}+x by 8.

$$8x^{2}+8x=96$$

Subtract 96 from both sides.

$$8x^{2}+8x-96=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 8 for a, 8 for b, and -96 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-8±\sqrt{8^{2}-4\times 8\left(-96\right)}}{2\times 8}$$

Square 8.

$$x=\frac{-8±\sqrt{64-4\times 8\left(-96\right)}}{2\times 8}$$

Multiply -4 times 8.

$$x=\frac{-8±\sqrt{64-32\left(-96\right)}}{2\times 8}$$

Multiply -32 times -96.

$$x=\frac{-8±\sqrt{64+3072}}{2\times 8}$$

Add 64 to 3072.

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

Take the square root of 3136.

$$x=\frac{-8±56}{2\times 8}$$

Multiply 2 times 8.

$$x=\frac{-8±56}{16}$$

Now solve the equation x=\frac{-8±56}{16} when ± is plus. Add -8 to 56.

$$x=\frac{48}{16}$$

Divide 48 by 16.

$$x=3$$

Now solve the equation x=\frac{-8±56}{16} when ± is minus. Subtract 56 from -8.

$$x=-\frac{64}{16}$$

Divide -64 by 16.

$$x=-4$$

The equation is now solved.

$$x=3 x=-4$$

Steps for Completing the Square

Use the distributive property to multiply x by x+1.

$$\left(x^{2}+x\right)\times 8=96$$

Use the distributive property to multiply x^{2}+x by 8.

$$8x^{2}+8x=96$$

Divide both sides by 8.

$$\frac{8x^{2}+8x}{8}=\frac{96}{8}$$

Dividing by 8 undoes the multiplication by 8.

$$x^{2}+\frac{8}{8}x=\frac{96}{8}$$

Divide 8 by 8.

$$x^{2}+x=\frac{96}{8}$$

Divide 96 by 8.

$$x^{2}+x=12$$

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

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

Square \frac{1}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+x+\frac{1}{4}=12+\frac{1}{4}$$

Add 12 to \frac{1}{4}.

$$x^{2}+x+\frac{1}{4}=\frac{49}{4}$$

Factor x^{2}+x+\frac{1}{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}.

$$\left(x+\frac{1}{2}\right)^{2}=\frac{49}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{1}{2}\right)^{2}}=\sqrt{\frac{49}{4}}$$

Simplify.

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

Subtract \frac{1}{2} from both sides of the equation.

$$x=3 x=-4$$