Solve x^2+x^2+1+2x-145=0

$${ x }^{ 2 } + { x }^{ 2 } +1+2x-145 = 0$$

Solve for x

$$x=-9$$

$$x=8$$

Steps Using Factoring By Grouping

Combine x^{2} and x^{2} to get 2x^{2}.

$$2x^{2}+1+2x-145=0$$

Subtract 145 from 1 to get -144.

$$2x^{2}-144+2x=0$$

Divide both sides by 2.

$$x^{2}-72+x=0$$

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

$$x^{2}+x-72=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx-72. To find a and b, set up a system to be solved.

$$a+b=1 ab=1\left(-72\right)=-72$$

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 -72.

$$-1,72 -2,36 -3,24 -4,18 -6,12 -8,9$$

Calculate the sum for each pair.

$$-1+72=71 -2+36=34 -3+24=21 -4+18=14 -6+12=6 -8+9=1$$

The solution is the pair that gives sum 1.

$$a=-8 b=9$$

Rewrite x^{2}+x-72 as \left(x^{2}-8x\right)+\left(9x-72\right).

$$\left(x^{2}-8x\right)+\left(9x-72\right)$$

Factor out x in the first and 9 in the second group.

$$x\left(x-8\right)+9\left(x-8\right)$$

Factor out common term x-8 by using distributive property.

$$\left(x-8\right)\left(x+9\right)$$

To find equation solutions, solve x-8=0 and x+9=0.

$$x=8 x=-9$$

Steps Using the Quadratic Formula

Combine x^{2} and x^{2} to get 2x^{2}.

$$2x^{2}+1+2x-145=0$$

Subtract 145 from 1 to get -144.

$$2x^{2}-144+2x=0$$

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.

$$2x^{2}+2x-144=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, 2 for b, and -144 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-2±\sqrt{2^{2}-4\times 2\left(-144\right)}}{2\times 2}$$

Square 2.

$$x=\frac{-2±\sqrt{4-4\times 2\left(-144\right)}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-2±\sqrt{4-8\left(-144\right)}}{2\times 2}$$

Multiply -8 times -144.

$$x=\frac{-2±\sqrt{4+1152}}{2\times 2}$$

Add 4 to 1152.

$$x=\frac{-2±\sqrt{1156}}{2\times 2}$$

Take the square root of 1156.

$$x=\frac{-2±34}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{-2±34}{4}$$

Now solve the equation x=\frac{-2±34}{4} when ± is plus. Add -2 to 34.

$$x=\frac{32}{4}$$

Divide 32 by 4.

$$x=8$$

Now solve the equation x=\frac{-2±34}{4} when ± is minus. Subtract 34 from -2.

$$x=-\frac{36}{4}$$

Divide -36 by 4.

$$x=-9$$

The equation is now solved.

$$x=8 x=-9$$

Steps for Completing the Square

Combine x^{2} and x^{2} to get 2x^{2}.

$$2x^{2}+1+2x-145=0$$

Subtract 145 from 1 to get -144.

$$2x^{2}-144+2x=0$$

Add 144 to both sides. Anything plus zero gives itself.

$$2x^{2}+2x=144$$

Divide both sides by 2.

$$\frac{2x^{2}+2x}{2}=\frac{144}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$x^{2}+\frac{2}{2}x=\frac{144}{2}$$

Divide 2 by 2.

$$x^{2}+x=\frac{144}{2}$$

Divide 144 by 2.

$$x^{2}+x=72$$

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.

$$x^{2}+x+\left(\frac{1}{2}\right)^{2}=72+\left(\frac{1}{2}\right)^{2}$$

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

$$x^{2}+x+\frac{1}{4}=72+\frac{1}{4}$$

Add 72 to \frac{1}{4}.

$$x^{2}+x+\frac{1}{4}=\frac{289}{4}$$

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}.

$$\left(x+\frac{1}{2}\right)^{2}=\frac{289}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{1}{2}\right)^{2}}=\sqrt{\frac{289}{4}}$$

Simplify.

$$x+\frac{1}{2}=\frac{17}{2} x+\frac{1}{2}=-\frac{17}{2}$$

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

$$x=8 x=-9$$