Solve 7x+left(x+1right)^2=5

$$7x+ { \left(x+1 \right) }^{ 2 } =5$$

Solve for x

$$x=\frac{\sqrt{97}-9}{2}\approx 0.424428901$$

$$x=\frac{-\sqrt{97}-9}{2}\approx -9.424428901$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+1\right)^{2}.

$$7x+x^{2}+2x+1=5$$

Combine 7x and 2x to get 9x.

$$9x+x^{2}+1=5$$

Subtract 5 from both sides.

$$9x+x^{2}+1-5=0$$

Subtract 5 from 1 to get -4.

$$9x+x^{2}-4=0$$

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.

$$x^{2}+9x-4=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 9 for b, and -4 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-9±\sqrt{9^{2}-4\left(-4\right)}}{2}$$

Square 9.

$$x=\frac{-9±\sqrt{81-4\left(-4\right)}}{2}$$

Multiply -4 times -4.

$$x=\frac{-9±\sqrt{81+16}}{2}$$

Add 81 to 16.

$$x=\frac{-9±\sqrt{97}}{2}$$

Now solve the equation x=\frac{-9±\sqrt{97}}{2} when ± is plus. Add -9 to \sqrt{97}.

$$x=\frac{\sqrt{97}-9}{2}$$

Now solve the equation x=\frac{-9±\sqrt{97}}{2} when ± is minus. Subtract \sqrt{97} from -9.

$$x=\frac{-\sqrt{97}-9}{2}$$

The equation is now solved.

$$x=\frac{\sqrt{97}-9}{2} x=\frac{-\sqrt{97}-9}{2}$$

Steps for Completing the Square

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+1\right)^{2}.

$$7x+x^{2}+2x+1=5$$

Combine 7x and 2x to get 9x.

$$9x+x^{2}+1=5$$

Subtract 1 from both sides.

$$9x+x^{2}=5-1$$

Subtract 1 from 5 to get 4.

$$9x+x^{2}=4$$

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.

$$x^{2}+9x=4$$

Divide 9, the coefficient of the x term, by 2 to get \frac{9}{2}. Then add the square of \frac{9}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+9x+\left(\frac{9}{2}\right)^{2}=4+\left(\frac{9}{2}\right)^{2}$$

Square \frac{9}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+9x+\frac{81}{4}=4+\frac{81}{4}$$

Add 4 to \frac{81}{4}.

$$x^{2}+9x+\frac{81}{4}=\frac{97}{4}$$

Factor x^{2}+9x+\frac{81}{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}.

$$\left(x+\frac{9}{2}\right)^{2}=\frac{97}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{9}{2}\right)^{2}}=\sqrt{\frac{97}{4}}$$

Simplify.

$$x+\frac{9}{2}=\frac{\sqrt{97}}{2} x+\frac{9}{2}=-\frac{\sqrt{97}}{2}$$

Subtract \frac{9}{2} from both sides of the equation.

$$x=\frac{\sqrt{97}-9}{2} x=\frac{-\sqrt{97}-9}{2}$$