Solve (x)(x+1)=812

$$( x ) ( x + 1 ) = 812$$

Solve for x

$$x=-29$$

$$x=28$$

Steps Using the Quadratic Formula

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

$$x^{2}+x=812$$

Subtract 812 from both sides.

$$x^{2}+x-812=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 1 for b, and -812 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-1±\sqrt{1^{2}-4\left(-812\right)}}{2}$$

Square 1.

$$x=\frac{-1±\sqrt{1-4\left(-812\right)}}{2}$$

Multiply -4 times -812.

$$x=\frac{-1±\sqrt{1+3248}}{2}$$

Add 1 to 3248.

$$x=\frac{-1±\sqrt{3249}}{2}$$

Take the square root of 3249.

$$x=\frac{-1±57}{2}$$

Now solve the equation x=\frac{-1±57}{2} when ± is plus. Add -1 to 57.

$$x=\frac{56}{2}$$

Divide 56 by 2.

$$x=28$$

Now solve the equation x=\frac{-1±57}{2} when ± is minus. Subtract 57 from -1.

$$x=-\frac{58}{2}$$

Divide -58 by 2.

$$x=-29$$

The equation is now solved.

$$x=28 x=-29$$

Steps for Completing the Square

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

$$x^{2}+x=812$$

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.

$$x^{2}+x+\left(\frac{1}{2}\right)^{2}=812+\left(\frac{1}{2}\right)^{2}$$

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

$$x^{2}+x+\frac{1}{4}=812+\frac{1}{4}$$

Add 812 to \frac{1}{4}.

$$x^{2}+x+\frac{1}{4}=\frac{3249}{4}$$

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}.

$$\left(x+\frac{1}{2}\right)^{2}=\frac{3249}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{1}{2}\right)^{2}}=\sqrt{\frac{3249}{4}}$$

Simplify.

$$x+\frac{1}{2}=\frac{57}{2} x+\frac{1}{2}=-\frac{57}{2}$$

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

$$x=28 x=-29$$