Solve -3/8x^2+9/4x+1/2=0quad-3x^2

$$- \frac { 3 } { 8 } x ^ { 2 } + \frac { 9 } { 4 } x + \frac { 1 } { 2 } = 0 \quad - 3 x ^ { 2 }$$

Solve for x (complex solution)

$$x=\frac{\sqrt{3}i}{21}-\frac{3}{7}\approx -0.428571429+0.08247861i$$

$$x=-\frac{\sqrt{3}i}{21}-\frac{3}{7}\approx -0.428571429-0.08247861i$$

Steps Using the Quadratic Formula

Anything plus zero gives itself.

$$-\frac{3}{8}x^{2}+\frac{9}{4}x+\frac{1}{2}=-3x^{2}$$

Add 3x^{2} to both sides.

$$-\frac{3}{8}x^{2}+\frac{9}{4}x+\frac{1}{2}+3x^{2}=0$$

Combine -\frac{3}{8}x^{2} and 3x^{2} to get \frac{21}{8}x^{2}.

$$\frac{21}{8}x^{2}+\frac{9}{4}x+\frac{1}{2}=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute \frac{21}{8} for a, \frac{9}{4} for b, and \frac{1}{2} for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\frac{9}{4}±\sqrt{\left(\frac{9}{4}\right)^{2}-4\times \frac{21}{8}\times \frac{1}{2}}}{2\times \frac{21}{8}}$$

Square \frac{9}{4} by squaring both the numerator and the denominator of the fraction.

$$x=\frac{-\frac{9}{4}±\sqrt{\frac{81}{16}-4\times \frac{21}{8}\times \frac{1}{2}}}{2\times \frac{21}{8}}$$

Multiply -4 times \frac{21}{8}.

$$x=\frac{-\frac{9}{4}±\sqrt{\frac{81}{16}-\frac{21}{2}\times \frac{1}{2}}}{2\times \frac{21}{8}}$$

Multiply -\frac{21}{2} times \frac{1}{2} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$x=\frac{-\frac{9}{4}±\sqrt{\frac{81}{16}-\frac{21}{4}}}{2\times \frac{21}{8}}$$

Add \frac{81}{16} to -\frac{21}{4} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x=\frac{-\frac{9}{4}±\sqrt{-\frac{3}{16}}}{2\times \frac{21}{8}}$$

Take the square root of -\frac{3}{16}.

$$x=\frac{-\frac{9}{4}±\frac{\sqrt{3}i}{4}}{2\times \frac{21}{8}}$$

Multiply 2 times \frac{21}{8}.

$$x=\frac{-\frac{9}{4}±\frac{\sqrt{3}i}{4}}{\frac{21}{4}}$$

Now solve the equation x=\frac{-\frac{9}{4}±\frac{\sqrt{3}i}{4}}{\frac{21}{4}} when ± is plus. Add -\frac{9}{4} to \frac{i\sqrt{3}}{4}.

$$x=\frac{-9+\sqrt{3}i}{4\times \frac{21}{4}}$$

Divide \frac{-9+i\sqrt{3}}{4} by \frac{21}{4} by multiplying \frac{-9+i\sqrt{3}}{4} by the reciprocal of \frac{21}{4}.

$$x=\frac{\sqrt{3}i}{21}-\frac{3}{7}$$

Now solve the equation x=\frac{-\frac{9}{4}±\frac{\sqrt{3}i}{4}}{\frac{21}{4}} when ± is minus. Subtract \frac{i\sqrt{3}}{4} from -\frac{9}{4}.

$$x=\frac{-\sqrt{3}i-9}{4\times \frac{21}{4}}$$

Divide \frac{-9-i\sqrt{3}}{4} by \frac{21}{4} by multiplying \frac{-9-i\sqrt{3}}{4} by the reciprocal of \frac{21}{4}.

$$x=-\frac{\sqrt{3}i}{21}-\frac{3}{7}$$

The equation is now solved.

$$x=\frac{\sqrt{3}i}{21}-\frac{3}{7} x=-\frac{\sqrt{3}i}{21}-\frac{3}{7}$$

Steps for Completing the Square

Anything plus zero gives itself.

$$-\frac{3}{8}x^{2}+\frac{9}{4}x+\frac{1}{2}=-3x^{2}$$

Add 3x^{2} to both sides.

$$-\frac{3}{8}x^{2}+\frac{9}{4}x+\frac{1}{2}+3x^{2}=0$$

Combine -\frac{3}{8}x^{2} and 3x^{2} to get \frac{21}{8}x^{2}.

$$\frac{21}{8}x^{2}+\frac{9}{4}x+\frac{1}{2}=0$$

Subtract \frac{1}{2} from both sides. Anything subtracted from zero gives its negation.

$$\frac{21}{8}x^{2}+\frac{9}{4}x=-\frac{1}{2}$$

Divide both sides of the equation by \frac{21}{8}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$\frac{\frac{21}{8}x^{2}+\frac{9}{4}x}{\frac{21}{8}}=-\frac{\frac{1}{2}}{\frac{21}{8}}$$

Dividing by \frac{21}{8} undoes the multiplication by \frac{21}{8}.

$$x^{2}+\frac{\frac{9}{4}}{\frac{21}{8}}x=-\frac{\frac{1}{2}}{\frac{21}{8}}$$

Divide \frac{9}{4} by \frac{21}{8} by multiplying \frac{9}{4} by the reciprocal of \frac{21}{8}.

$$x^{2}+\frac{6}{7}x=-\frac{\frac{1}{2}}{\frac{21}{8}}$$

Divide -\frac{1}{2} by \frac{21}{8} by multiplying -\frac{1}{2} by the reciprocal of \frac{21}{8}.

$$x^{2}+\frac{6}{7}x=-\frac{4}{21}$$

Divide \frac{6}{7}, the coefficient of the x term, by 2 to get \frac{3}{7}. Then add the square of \frac{3}{7} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{6}{7}x+\left(\frac{3}{7}\right)^{2}=-\frac{4}{21}+\left(\frac{3}{7}\right)^{2}$$

Square \frac{3}{7} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{6}{7}x+\frac{9}{49}=-\frac{4}{21}+\frac{9}{49}$$

Add -\frac{4}{21} to \frac{9}{49} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{6}{7}x+\frac{9}{49}=-\frac{1}{147}$$

Factor x^{2}+\frac{6}{7}x+\frac{9}{49}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{3}{7}\right)^{2}=-\frac{1}{147}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{3}{7}\right)^{2}}=\sqrt{-\frac{1}{147}}$$

Simplify.

$$x+\frac{3}{7}=\frac{\sqrt{3}i}{21} x+\frac{3}{7}=-\frac{\sqrt{3}i}{21}$$

Subtract \frac{3}{7} from both sides of the equation.

$$x=\frac{\sqrt{3}i}{21}-\frac{3}{7} x=-\frac{\sqrt{3}i}{21}-\frac{3}{7}$$