Solve 1/2x^2-8=3x

$$\frac { 1 } { 2 } x ^ { 2 } - 8 = 3 x$$

Solve for x

$$x=-2$$

$$x=8$$

Steps Using the Quadratic Formula

Subtract 3x from both sides.

$$\frac{1}{2}x^{2}-8-3x=0$$

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.

$$\frac{1}{2}x^{2}-3x-8=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute \frac{1}{2} for a, -3 for b, and -8 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-3\right)±\sqrt{\left(-3\right)^{2}-4\times \frac{1}{2}\left(-8\right)}}{2\times \frac{1}{2}}$$

Square -3.

$$x=\frac{-\left(-3\right)±\sqrt{9-4\times \frac{1}{2}\left(-8\right)}}{2\times \frac{1}{2}}$$

Multiply -4 times \frac{1}{2}.

$$x=\frac{-\left(-3\right)±\sqrt{9-2\left(-8\right)}}{2\times \frac{1}{2}}$$

Multiply -2 times -8.

$$x=\frac{-\left(-3\right)±\sqrt{9+16}}{2\times \frac{1}{2}}$$

Add 9 to 16.

$$x=\frac{-\left(-3\right)±\sqrt{25}}{2\times \frac{1}{2}}$$

Take the square root of 25.

$$x=\frac{-\left(-3\right)±5}{2\times \frac{1}{2}}$$

The opposite of -3 is 3.

$$x=\frac{3±5}{2\times \frac{1}{2}}$$

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

$$x=\frac{3±5}{1}$$

Now solve the equation x=\frac{3±5}{1} when ± is plus. Add 3 to 5.

$$x=\frac{8}{1}$$

Divide 8 by 1.

$$x=8$$

Now solve the equation x=\frac{3±5}{1} when ± is minus. Subtract 5 from 3.

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

Divide -2 by 1.

$$x=-2$$

The equation is now solved.

$$x=8 x=-2$$

Steps for Completing the Square

Subtract 3x from both sides.

$$\frac{1}{2}x^{2}-8-3x=0$$

Add 8 to both sides. Anything plus zero gives itself.

$$\frac{1}{2}x^{2}-3x=8$$

Multiply both sides by 2.

$$\frac{\frac{1}{2}x^{2}-3x}{\frac{1}{2}}=\frac{8}{\frac{1}{2}}$$

Dividing by \frac{1}{2} undoes the multiplication by \frac{1}{2}.

$$x^{2}+\left(-\frac{3}{\frac{1}{2}}\right)x=\frac{8}{\frac{1}{2}}$$

Divide -3 by \frac{1}{2} by multiplying -3 by the reciprocal of \frac{1}{2}.

$$x^{2}-6x=\frac{8}{\frac{1}{2}}$$

Divide 8 by \frac{1}{2} by multiplying 8 by the reciprocal of \frac{1}{2}.

$$x^{2}-6x=16$$

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

$$x^{2}-6x+\left(-3\right)^{2}=16+\left(-3\right)^{2}$$

Square -3.

$$x^{2}-6x+9=16+9$$

Add 16 to 9.

$$x^{2}-6x+9=25$$

Factor x^{2}-6x+9. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-3\right)^{2}=25$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-3\right)^{2}}=\sqrt{25}$$

Simplify.

$$x-3=5 x-3=-5$$

Add 3 to both sides of the equation.

$$x=8 x=-2$$