Solve left(x-11right)^2+{left(3/4x-7+5right)}^2=100

$${ \left(x-11 \right) }^{ 2 } + { \left( \frac{ 3 }{ 4 } x-7+5 \right) }^{ 2 } =100$$

Solve for x

$$x=4\sqrt{3}+8\approx 14.92820323$$

$$x=8-4\sqrt{3}\approx 1.07179677$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-11\right)^{2}.

$$x^{2}-22x+121+\left(\frac{3}{4}x-7+5\right)^{2}=100$$

Add -7 and 5 to get -2.

$$x^{2}-22x+121+\left(\frac{3}{4}x-2\right)^{2}=100$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(\frac{3}{4}x-2\right)^{2}.

$$x^{2}-22x+121+\frac{9}{16}x^{2}-3x+4=100$$

Combine x^{2} and \frac{9}{16}x^{2} to get \frac{25}{16}x^{2}.

$$\frac{25}{16}x^{2}-22x+121-3x+4=100$$

Combine -22x and -3x to get -25x.

$$\frac{25}{16}x^{2}-25x+121+4=100$$

Add 121 and 4 to get 125.

$$\frac{25}{16}x^{2}-25x+125=100$$

Subtract 100 from both sides.

$$\frac{25}{16}x^{2}-25x+125-100=0$$

Subtract 100 from 125 to get 25.

$$\frac{25}{16}x^{2}-25x+25=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute \frac{25}{16} for a, -25 for b, and 25 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-25\right)±\sqrt{\left(-25\right)^{2}-4\times \frac{25}{16}\times 25}}{2\times \frac{25}{16}}$$

Square -25.

$$x=\frac{-\left(-25\right)±\sqrt{625-4\times \frac{25}{16}\times 25}}{2\times \frac{25}{16}}$$

Multiply -4 times \frac{25}{16}.

$$x=\frac{-\left(-25\right)±\sqrt{625-\frac{25}{4}\times 25}}{2\times \frac{25}{16}}$$

Multiply -\frac{25}{4} times 25.

$$x=\frac{-\left(-25\right)±\sqrt{625-\frac{625}{4}}}{2\times \frac{25}{16}}$$

Add 625 to -\frac{625}{4}.

$$x=\frac{-\left(-25\right)±\sqrt{\frac{1875}{4}}}{2\times \frac{25}{16}}$$

Take the square root of \frac{1875}{4}.

$$x=\frac{-\left(-25\right)±\frac{25\sqrt{3}}{2}}{2\times \frac{25}{16}}$$

The opposite of -25 is 25.

$$x=\frac{25±\frac{25\sqrt{3}}{2}}{2\times \frac{25}{16}}$$

Multiply 2 times \frac{25}{16}.

$$x=\frac{25±\frac{25\sqrt{3}}{2}}{\frac{25}{8}}$$

Now solve the equation x=\frac{25±\frac{25\sqrt{3}}{2}}{\frac{25}{8}} when ± is plus. Add 25 to \frac{25\sqrt{3}}{2}.

$$x=\frac{\frac{25\sqrt{3}}{2}+25}{\frac{25}{8}}$$

Divide 25+\frac{25\sqrt{3}}{2} by \frac{25}{8} by multiplying 25+\frac{25\sqrt{3}}{2} by the reciprocal of \frac{25}{8}.

$$x=4\sqrt{3}+8$$

Now solve the equation x=\frac{25±\frac{25\sqrt{3}}{2}}{\frac{25}{8}} when ± is minus. Subtract \frac{25\sqrt{3}}{2} from 25.

$$x=\frac{-\frac{25\sqrt{3}}{2}+25}{\frac{25}{8}}$$

Divide 25-\frac{25\sqrt{3}}{2} by \frac{25}{8} by multiplying 25-\frac{25\sqrt{3}}{2} by the reciprocal of \frac{25}{8}.

$$x=8-4\sqrt{3}$$

The equation is now solved.

$$x=4\sqrt{3}+8 x=8-4\sqrt{3}$$

Steps for Completing the Square

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-11\right)^{2}.

$$x^{2}-22x+121+\left(\frac{3}{4}x-7+5\right)^{2}=100$$

Add -7 and 5 to get -2.

$$x^{2}-22x+121+\left(\frac{3}{4}x-2\right)^{2}=100$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(\frac{3}{4}x-2\right)^{2}.

$$x^{2}-22x+121+\frac{9}{16}x^{2}-3x+4=100$$

Combine x^{2} and \frac{9}{16}x^{2} to get \frac{25}{16}x^{2}.

$$\frac{25}{16}x^{2}-22x+121-3x+4=100$$

Combine -22x and -3x to get -25x.

$$\frac{25}{16}x^{2}-25x+121+4=100$$

Add 121 and 4 to get 125.

$$\frac{25}{16}x^{2}-25x+125=100$$

Subtract 125 from both sides.

$$\frac{25}{16}x^{2}-25x=100-125$$

Subtract 125 from 100 to get -25.

$$\frac{25}{16}x^{2}-25x=-25$$

Divide both sides of the equation by \frac{25}{16}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$\frac{\frac{25}{16}x^{2}-25x}{\frac{25}{16}}=-\frac{25}{\frac{25}{16}}$$

Dividing by \frac{25}{16} undoes the multiplication by \frac{25}{16}.

$$x^{2}+\left(-\frac{25}{\frac{25}{16}}\right)x=-\frac{25}{\frac{25}{16}}$$

Divide -25 by \frac{25}{16} by multiplying -25 by the reciprocal of \frac{25}{16}.

$$x^{2}-16x=-\frac{25}{\frac{25}{16}}$$

Divide -25 by \frac{25}{16} by multiplying -25 by the reciprocal of \frac{25}{16}.

$$x^{2}-16x=-16$$

Divide -16, the coefficient of the x term, by 2 to get -8. Then add the square of -8 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-16x+\left(-8\right)^{2}=-16+\left(-8\right)^{2}$$

Square -8.

$$x^{2}-16x+64=-16+64$$

Add -16 to 64.

$$x^{2}-16x+64=48$$

Factor x^{2}-16x+64. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-8\right)^{2}=48$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-8\right)^{2}}=\sqrt{48}$$

Simplify.

$$x-8=4\sqrt{3} x-8=-4\sqrt{3}$$

Add 8 to both sides of the equation.

$$x=4\sqrt{3}+8 x=8-4\sqrt{3}$$