Solve x^2+(7x-25)^2/5-5x+3(7x-25/5)=10

$$x ^ { 2 } + \frac { ( 7 x - 25 ) ^ { 2 } } { 5 } - 5 x + 3 ( \frac { 7 x - 25 } { 5 } ) = 10$$

Solve for x

$$x = \frac{\sqrt{481} + 59}{18} \approx 4.496206233$$

$$x = \frac{59 - \sqrt{481}}{18} \approx 2.059349322$$

Steps Using the Quadratic Formula

Multiply both sides of the equation by 5.

$$5x^{2}+\left(7x-25\right)^{2}-25x+3\left(7x-25\right)=50$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(7x-25\right)^{2}.

$$5x^{2}+49x^{2}-350x+625-25x+3\left(7x-25\right)=50$$

Combine 5x^{2} and 49x^{2} to get 54x^{2}.

$$54x^{2}-350x+625-25x+3\left(7x-25\right)=50$$

Combine -350x and -25x to get -375x.

$$54x^{2}-375x+625+3\left(7x-25\right)=50$$

Use the distributive property to multiply 3 by 7x-25.

$$54x^{2}-375x+625+21x-75=50$$

Combine -375x and 21x to get -354x.

$$54x^{2}-354x+625-75=50$$

Subtract 75 from 625 to get 550.

$$54x^{2}-354x+550=50$$

Subtract 50 from both sides.

$$54x^{2}-354x+550-50=0$$

Subtract 50 from 550 to get 500.

$$54x^{2}-354x+500=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 54 for a, -354 for b, and 500 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-354\right)±\sqrt{\left(-354\right)^{2}-4\times 54\times 500}}{2\times 54}$$

Square -354.

$$x=\frac{-\left(-354\right)±\sqrt{125316-4\times 54\times 500}}{2\times 54}$$

Multiply -4 times 54.

$$x=\frac{-\left(-354\right)±\sqrt{125316-216\times 500}}{2\times 54}$$

Multiply -216 times 500.

$$x=\frac{-\left(-354\right)±\sqrt{125316-108000}}{2\times 54}$$

Add 125316 to -108000.

$$x=\frac{-\left(-354\right)±\sqrt{17316}}{2\times 54}$$

Take the square root of 17316.

$$x=\frac{-\left(-354\right)±6\sqrt{481}}{2\times 54}$$

The opposite of -354 is 354.

$$x=\frac{354±6\sqrt{481}}{2\times 54}$$

Multiply 2 times 54.

$$x=\frac{354±6\sqrt{481}}{108}$$

Now solve the equation x=\frac{354±6\sqrt{481}}{108} when ± is plus. Add 354 to 6\sqrt{481}.

$$x=\frac{6\sqrt{481}+354}{108}$$

Divide 354+6\sqrt{481} by 108.

$$x=\frac{\sqrt{481}+59}{18}$$

Now solve the equation x=\frac{354±6\sqrt{481}}{108} when ± is minus. Subtract 6\sqrt{481} from 354.

$$x=\frac{354-6\sqrt{481}}{108}$$

Divide 354-6\sqrt{481} by 108.

$$x=\frac{59-\sqrt{481}}{18}$$

The equation is now solved.

$$x=\frac{\sqrt{481}+59}{18} x=\frac{59-\sqrt{481}}{18}$$

Steps for Completing the Square

Multiply both sides of the equation by 5.

$$5x^{2}+\left(7x-25\right)^{2}-25x+3\left(7x-25\right)=50$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(7x-25\right)^{2}.

$$5x^{2}+49x^{2}-350x+625-25x+3\left(7x-25\right)=50$$

Combine 5x^{2} and 49x^{2} to get 54x^{2}.

$$54x^{2}-350x+625-25x+3\left(7x-25\right)=50$$

Combine -350x and -25x to get -375x.

$$54x^{2}-375x+625+3\left(7x-25\right)=50$$

Use the distributive property to multiply 3 by 7x-25.

$$54x^{2}-375x+625+21x-75=50$$

Combine -375x and 21x to get -354x.

$$54x^{2}-354x+625-75=50$$

Subtract 75 from 625 to get 550.

$$54x^{2}-354x+550=50$$

Subtract 550 from both sides.

$$54x^{2}-354x=50-550$$

Subtract 550 from 50 to get -500.

$$54x^{2}-354x=-500$$

Divide both sides by 54.

$$\frac{54x^{2}-354x}{54}=-\frac{500}{54}$$

Dividing by 54 undoes the multiplication by 54.

$$x^{2}+\left(-\frac{354}{54}\right)x=-\frac{500}{54}$$

Reduce the fraction \frac{-354}{54} to lowest terms by extracting and canceling out 6.

$$x^{2}-\frac{59}{9}x=-\frac{500}{54}$$

Reduce the fraction \frac{-500}{54} to lowest terms by extracting and canceling out 2.

$$x^{2}-\frac{59}{9}x=-\frac{250}{27}$$

Divide -\frac{59}{9}, the coefficient of the x term, by 2 to get -\frac{59}{18}. Then add the square of -\frac{59}{18} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{59}{9}x+\left(-\frac{59}{18}\right)^{2}=-\frac{250}{27}+\left(-\frac{59}{18}\right)^{2}$$

Square -\frac{59}{18} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{59}{9}x+\frac{3481}{324}=-\frac{250}{27}+\frac{3481}{324}$$

Add -\frac{250}{27} to \frac{3481}{324} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{59}{9}x+\frac{3481}{324}=\frac{481}{324}$$

Factor x^{2}-\frac{59}{9}x+\frac{3481}{324}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{59}{18}\right)^{2}=\frac{481}{324}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{59}{18}\right)^{2}}=\sqrt{\frac{481}{324}}$$

Simplify.

$$x-\frac{59}{18}=\frac{\sqrt{481}}{18} x-\frac{59}{18}=-\frac{\sqrt{481}}{18}$$

Add \frac{59}{18} to both sides of the equation.

$$x=\frac{\sqrt{481}+59}{18} x=\frac{59-\sqrt{481}}{18}$$