Solve 1/8x^2-5x+100=58

$$\frac{ 1 }{ 8 } { x }^{ 2 } -5x+100=58$$

Solve for x

$$x=12$$

$$x=28$$

Steps Using the Quadratic Formula

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.

$$\frac{1}{8}x^{2}-5x+100=58$$

Subtract 58 from both sides of the equation.

$$\frac{1}{8}x^{2}-5x+100-58=58-58$$

Subtracting 58 from itself leaves 0.

$$\frac{1}{8}x^{2}-5x+100-58=0$$

Subtract 58 from 100.

$$\frac{1}{8}x^{2}-5x+42=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute \frac{1}{8} for a, -5 for b, and 42 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-5\right)±\sqrt{\left(-5\right)^{2}-4\times \frac{1}{8}\times 42}}{2\times \frac{1}{8}}$$

Square -5.

$$x=\frac{-\left(-5\right)±\sqrt{25-4\times \frac{1}{8}\times 42}}{2\times \frac{1}{8}}$$

Multiply -4 times \frac{1}{8}.

$$x=\frac{-\left(-5\right)±\sqrt{25-\frac{1}{2}\times 42}}{2\times \frac{1}{8}}$$

Multiply -\frac{1}{2} times 42.

$$x=\frac{-\left(-5\right)±\sqrt{25-21}}{2\times \frac{1}{8}}$$

Add 25 to -21.

$$x=\frac{-\left(-5\right)±\sqrt{4}}{2\times \frac{1}{8}}$$

Take the square root of 4.

$$x=\frac{-\left(-5\right)±2}{2\times \frac{1}{8}}$$

The opposite of -5 is 5.

$$x=\frac{5±2}{2\times \frac{1}{8}}$$

Multiply 2 times \frac{1}{8}.

$$x=\frac{5±2}{\frac{1}{4}}$$

Now solve the equation x=\frac{5±2}{\frac{1}{4}} when ± is plus. Add 5 to 2.

$$x=\frac{7}{\frac{1}{4}}$$

Divide 7 by \frac{1}{4} by multiplying 7 by the reciprocal of \frac{1}{4}.

$$x=28$$

Now solve the equation x=\frac{5±2}{\frac{1}{4}} when ± is minus. Subtract 2 from 5.

$$x=\frac{3}{\frac{1}{4}}$$

Divide 3 by \frac{1}{4} by multiplying 3 by the reciprocal of \frac{1}{4}.

$$x=12$$

The equation is now solved.

$$x=28 x=12$$

Steps for Completing the Square

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.

$$\frac{1}{8}x^{2}-5x+100=58$$

Subtract 100 from both sides of the equation.

$$\frac{1}{8}x^{2}-5x+100-100=58-100$$

Subtracting 100 from itself leaves 0.

$$\frac{1}{8}x^{2}-5x=58-100$$

Subtract 100 from 58.

$$\frac{1}{8}x^{2}-5x=-42$$

Multiply both sides by 8.

$$\frac{\frac{1}{8}x^{2}-5x}{\frac{1}{8}}=-\frac{42}{\frac{1}{8}}$$

Dividing by \frac{1}{8} undoes the multiplication by \frac{1}{8}.

$$x^{2}+\left(-\frac{5}{\frac{1}{8}}\right)x=-\frac{42}{\frac{1}{8}}$$

Divide -5 by \frac{1}{8} by multiplying -5 by the reciprocal of \frac{1}{8}.

$$x^{2}-40x=-\frac{42}{\frac{1}{8}}$$

Divide -42 by \frac{1}{8} by multiplying -42 by the reciprocal of \frac{1}{8}.

$$x^{2}-40x=-336$$

Divide -40, the coefficient of the x term, by 2 to get -20. Then add the square of -20 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-40x+\left(-20\right)^{2}=-336+\left(-20\right)^{2}$$

Square -20.

$$x^{2}-40x+400=-336+400$$

Add -336 to 400.

$$x^{2}-40x+400=64$$

Factor x^{2}-40x+400. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-20\right)^{2}=64$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-20\right)^{2}}=\sqrt{64}$$

Simplify.

$$x-20=8 x-20=-8$$

Add 20 to both sides of the equation.

$$x=28 x=12$$