Solve 0=1/2x^2+3x-4

$$0 = \frac { 1 } { 2 } x ^ { 2 } + 3 x - 4$$

Solve for x (complex solution)

$$x=\sqrt{17}-3\approx 1.123105626$$

$$x=-\left(\sqrt{17}+3\right)\approx -7.123105626$$

Steps Using the Quadratic Formula

Swap sides so that all variable terms are on the left hand side.

$$\frac{1}{2}x^{2}+3x-4=0$$

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

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

Square 3.

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

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

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

Multiply -2 times -4.

$$x=\frac{-3±\sqrt{9+8}}{2\times \frac{1}{2}}$$

Add 9 to 8.

$$x=\frac{-3±\sqrt{17}}{2\times \frac{1}{2}}$$

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

$$x=\frac{-3±\sqrt{17}}{1}$$

Now solve the equation x=\frac{-3±\sqrt{17}}{1} when ± is plus. Add -3 to \sqrt{17}.

$$x=\frac{\sqrt{17}-3}{1}$$

Divide -3+\sqrt{17} by 1.

$$x=\sqrt{17}-3$$

Now solve the equation x=\frac{-3±\sqrt{17}}{1} when ± is minus. Subtract \sqrt{17} from -3.

$$x=\frac{-\sqrt{17}-3}{1}$$

Divide -3-\sqrt{17} by 1.

$$x=-\sqrt{17}-3$$

The equation is now solved.

$$x=\sqrt{17}-3 x=-\sqrt{17}-3$$

Steps for Completing the Square

Swap sides so that all variable terms are on the left hand side.

$$\frac{1}{2}x^{2}+3x-4=0$$

Add 4 to both sides. Anything plus zero gives itself.

$$\frac{1}{2}x^{2}+3x=4$$

Multiply both sides by 2.

$$\frac{\frac{1}{2}x^{2}+3x}{\frac{1}{2}}=\frac{4}{\frac{1}{2}}$$

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

$$x^{2}+\frac{3}{\frac{1}{2}}x=\frac{4}{\frac{1}{2}}$$

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

$$x^{2}+6x=\frac{4}{\frac{1}{2}}$$

Divide 4 by \frac{1}{2} by multiplying 4 by the reciprocal of \frac{1}{2}.

$$x^{2}+6x=8$$

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.

$$x^{2}+6x+3^{2}=8+3^{2}$$

Square 3.

$$x^{2}+6x+9=8+9$$

Add 8 to 9.

$$x^{2}+6x+9=17$$

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}.

$$\left(x+3\right)^{2}=17$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+3\right)^{2}}=\sqrt{17}$$

Simplify.

$$x+3=\sqrt{17} x+3=-\sqrt{17}$$

Subtract 3 from both sides of the equation.

$$x=\sqrt{17}-3 x=-\sqrt{17}-3$$

Solve for x

$$x=\sqrt{17}-3\approx 1.123105626$$

$$x=-\sqrt{17}-3\approx -7.123105626$$

Steps Using the Quadratic Formula

Swap sides so that all variable terms are on the left hand side.

$$\frac{1}{2}x^{2}+3x-4=0$$

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

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

Square 3.

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

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

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

Multiply -2 times -4.

$$x=\frac{-3±\sqrt{9+8}}{2\times \frac{1}{2}}$$

Add 9 to 8.

$$x=\frac{-3±\sqrt{17}}{2\times \frac{1}{2}}$$

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

$$x=\frac{-3±\sqrt{17}}{1}$$

Now solve the equation x=\frac{-3±\sqrt{17}}{1} when ± is plus. Add -3 to \sqrt{17}.

$$x=\frac{\sqrt{17}-3}{1}$$

Divide -3+\sqrt{17} by 1.

$$x=\sqrt{17}-3$$

Now solve the equation x=\frac{-3±\sqrt{17}}{1} when ± is minus. Subtract \sqrt{17} from -3.

$$x=\frac{-\sqrt{17}-3}{1}$$

Divide -3-\sqrt{17} by 1.

$$x=-\sqrt{17}-3$$

The equation is now solved.

$$x=\sqrt{17}-3 x=-\sqrt{17}-3$$

Steps for Completing the Square

Swap sides so that all variable terms are on the left hand side.

$$\frac{1}{2}x^{2}+3x-4=0$$

Add 4 to both sides. Anything plus zero gives itself.

$$\frac{1}{2}x^{2}+3x=4$$

Multiply both sides by 2.

$$\frac{\frac{1}{2}x^{2}+3x}{\frac{1}{2}}=\frac{4}{\frac{1}{2}}$$

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

$$x^{2}+\frac{3}{\frac{1}{2}}x=\frac{4}{\frac{1}{2}}$$

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

$$x^{2}+6x=\frac{4}{\frac{1}{2}}$$

Divide 4 by \frac{1}{2} by multiplying 4 by the reciprocal of \frac{1}{2}.

$$x^{2}+6x=8$$

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.

$$x^{2}+6x+3^{2}=8+3^{2}$$

Square 3.

$$x^{2}+6x+9=8+9$$

Add 8 to 9.

$$x^{2}+6x+9=17$$

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}.

$$\left(x+3\right)^{2}=17$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+3\right)^{2}}=\sqrt{17}$$

Simplify.

$$x+3=\sqrt{17} x+3=-\sqrt{17}$$

Subtract 3 from both sides of the equation.

$$x=\sqrt{17}-3 x=-\sqrt{17}-3$$