Solve {l}{text{Losop.}}{text{a}(2x-5)^2-(x+1)^2=7x}

$$\left. \begin{array} { l } { \text { Los op. } } \\ { \text { a } ( 2 x - 5 ) ^ { 2 } - ( x + 1 ) ^ { 2 } = 7 x } \end{array} \right.$$

Solve for x

$$x = \frac{\sqrt{553} + 29}{6} \approx 8.752658672$$

$$x=\frac{29-\sqrt{553}}{6}\approx 0.914007995$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(2x-5\right)^{2}.

$$4x^{2}-20x+25-\left(x+1\right)^{2}=7x$$

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

$$4x^{2}-20x+25-\left(x^{2}+2x+1\right)=7x$$

To find the opposite of x^{2}+2x+1, find the opposite of each term.

$$4x^{2}-20x+25-x^{2}-2x-1=7x$$

Combine 4x^{2} and -x^{2} to get 3x^{2}.

$$3x^{2}-20x+25-2x-1=7x$$

Combine -20x and -2x to get -22x.

$$3x^{2}-22x+25-1=7x$$

Subtract 1 from 25 to get 24.

$$3x^{2}-22x+24=7x$$

Subtract 7x from both sides.

$$3x^{2}-22x+24-7x=0$$

Combine -22x and -7x to get -29x.

$$3x^{2}-29x+24=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, -29 for b, and 24 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-29\right)±\sqrt{\left(-29\right)^{2}-4\times 3\times 24}}{2\times 3}$$

Square -29.

$$x=\frac{-\left(-29\right)±\sqrt{841-4\times 3\times 24}}{2\times 3}$$

Multiply -4 times 3.

$$x=\frac{-\left(-29\right)±\sqrt{841-12\times 24}}{2\times 3}$$

Multiply -12 times 24.

$$x=\frac{-\left(-29\right)±\sqrt{841-288}}{2\times 3}$$

Add 841 to -288.

$$x=\frac{-\left(-29\right)±\sqrt{553}}{2\times 3}$$

The opposite of -29 is 29.

$$x=\frac{29±\sqrt{553}}{2\times 3}$$

Multiply 2 times 3.

$$x=\frac{29±\sqrt{553}}{6}$$

Now solve the equation x=\frac{29±\sqrt{553}}{6} when ± is plus. Add 29 to \sqrt{553}.

$$x=\frac{\sqrt{553}+29}{6}$$

Now solve the equation x=\frac{29±\sqrt{553}}{6} when ± is minus. Subtract \sqrt{553} from 29.

$$x=\frac{29-\sqrt{553}}{6}$$

The equation is now solved.

$$x=\frac{\sqrt{553}+29}{6} x=\frac{29-\sqrt{553}}{6}$$

Steps for Completing the Square

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(2x-5\right)^{2}.

$$4x^{2}-20x+25-\left(x+1\right)^{2}=7x$$

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

$$4x^{2}-20x+25-\left(x^{2}+2x+1\right)=7x$$

To find the opposite of x^{2}+2x+1, find the opposite of each term.

$$4x^{2}-20x+25-x^{2}-2x-1=7x$$

Combine 4x^{2} and -x^{2} to get 3x^{2}.

$$3x^{2}-20x+25-2x-1=7x$$

Combine -20x and -2x to get -22x.

$$3x^{2}-22x+25-1=7x$$

Subtract 1 from 25 to get 24.

$$3x^{2}-22x+24=7x$$

Subtract 7x from both sides.

$$3x^{2}-22x+24-7x=0$$

Combine -22x and -7x to get -29x.

$$3x^{2}-29x+24=0$$

Subtract 24 from both sides. Anything subtracted from zero gives its negation.

$$3x^{2}-29x=-24$$

Divide both sides by 3.

$$\frac{3x^{2}-29x}{3}=-\frac{24}{3}$$

Dividing by 3 undoes the multiplication by 3.

$$x^{2}-\frac{29}{3}x=-\frac{24}{3}$$

Divide -24 by 3.

$$x^{2}-\frac{29}{3}x=-8$$

Divide -\frac{29}{3}, the coefficient of the x term, by 2 to get -\frac{29}{6}. Then add the square of -\frac{29}{6} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{29}{3}x+\left(-\frac{29}{6}\right)^{2}=-8+\left(-\frac{29}{6}\right)^{2}$$

Square -\frac{29}{6} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{29}{3}x+\frac{841}{36}=-8+\frac{841}{36}$$

Add -8 to \frac{841}{36}.

$$x^{2}-\frac{29}{3}x+\frac{841}{36}=\frac{553}{36}$$

Factor x^{2}-\frac{29}{3}x+\frac{841}{36}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{29}{6}\right)^{2}=\frac{553}{36}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{29}{6}\right)^{2}}=\sqrt{\frac{553}{36}}$$

Simplify.

$$x-\frac{29}{6}=\frac{\sqrt{553}}{6} x-\frac{29}{6}=-\frac{\sqrt{553}}{6}$$

Add \frac{29}{6} to both sides of the equation.

$$x=\frac{\sqrt{553}+29}{6} x=\frac{29-\sqrt{553}}{6}$$