Solve 6=-(m-1+3/2m)^2+3/4m^2

$$6 = - ( m - 1 + \frac { 3 } { 2 } m ) ^ { 2 } + \frac { 3 } { 4 } m ^ { 2 }$$

Solve for m

$$m=\frac{5+\sqrt{129}i}{11}\approx 0.454545455+1.03252879i$$

$$m=\frac{-\sqrt{129}i+5}{11}\approx 0.454545455-1.03252879i$$

Steps Using the Quadratic Formula

Combine m and \frac{3}{2}m to get \frac{5}{2}m.

$$6=-\left(\frac{5}{2}m-1\right)^{2}+\frac{3}{4}m^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(\frac{5}{2}m-1\right)^{2}.

$$6=-\left(\frac{25}{4}m^{2}-5m+1\right)+\frac{3}{4}m^{2}$$

To find the opposite of \frac{25}{4}m^{2}-5m+1, find the opposite of each term.

$$6=-\frac{25}{4}m^{2}+5m-1+\frac{3}{4}m^{2}$$

Combine -\frac{25}{4}m^{2} and \frac{3}{4}m^{2} to get -\frac{11}{2}m^{2}.

$$6=-\frac{11}{2}m^{2}+5m-1$$

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

$$-\frac{11}{2}m^{2}+5m-1=6$$

Subtract 6 from both sides.

$$-\frac{11}{2}m^{2}+5m-1-6=0$$

Subtract 6 from -1 to get -7.

$$-\frac{11}{2}m^{2}+5m-7=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -\frac{11}{2} for a, 5 for b, and -7 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$m=\frac{-5±\sqrt{5^{2}-4\left(-\frac{11}{2}\right)\left(-7\right)}}{2\left(-\frac{11}{2}\right)}$$

Square 5.

$$m=\frac{-5±\sqrt{25-4\left(-\frac{11}{2}\right)\left(-7\right)}}{2\left(-\frac{11}{2}\right)}$$

Multiply -4 times -\frac{11}{2}.

$$m=\frac{-5±\sqrt{25+22\left(-7\right)}}{2\left(-\frac{11}{2}\right)}$$

Multiply 22 times -7.

$$m=\frac{-5±\sqrt{25-154}}{2\left(-\frac{11}{2}\right)}$$

Add 25 to -154.

$$m=\frac{-5±\sqrt{-129}}{2\left(-\frac{11}{2}\right)}$$

Take the square root of -129.

$$m=\frac{-5±\sqrt{129}i}{2\left(-\frac{11}{2}\right)}$$

Multiply 2 times -\frac{11}{2}.

$$m=\frac{-5±\sqrt{129}i}{-11}$$

Now solve the equation m=\frac{-5±\sqrt{129}i}{-11} when ± is plus. Add -5 to i\sqrt{129}.

$$m=\frac{-5+\sqrt{129}i}{-11}$$

Divide -5+i\sqrt{129} by -11.

$$m=\frac{-\sqrt{129}i+5}{11}$$

Now solve the equation m=\frac{-5±\sqrt{129}i}{-11} when ± is minus. Subtract i\sqrt{129} from -5.

$$m=\frac{-\sqrt{129}i-5}{-11}$$

Divide -5-i\sqrt{129} by -11.

$$m=\frac{5+\sqrt{129}i}{11}$$

The equation is now solved.

$$m=\frac{-\sqrt{129}i+5}{11} m=\frac{5+\sqrt{129}i}{11}$$

Steps for Completing the Square

Combine m and \frac{3}{2}m to get \frac{5}{2}m.

$$6=-\left(\frac{5}{2}m-1\right)^{2}+\frac{3}{4}m^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(\frac{5}{2}m-1\right)^{2}.

$$6=-\left(\frac{25}{4}m^{2}-5m+1\right)+\frac{3}{4}m^{2}$$

To find the opposite of \frac{25}{4}m^{2}-5m+1, find the opposite of each term.

$$6=-\frac{25}{4}m^{2}+5m-1+\frac{3}{4}m^{2}$$

Combine -\frac{25}{4}m^{2} and \frac{3}{4}m^{2} to get -\frac{11}{2}m^{2}.

$$6=-\frac{11}{2}m^{2}+5m-1$$

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

$$-\frac{11}{2}m^{2}+5m-1=6$$

Add 1 to both sides.

$$-\frac{11}{2}m^{2}+5m=6+1$$

Add 6 and 1 to get 7.

$$-\frac{11}{2}m^{2}+5m=7$$

Divide both sides of the equation by -\frac{11}{2}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$\frac{-\frac{11}{2}m^{2}+5m}{-\frac{11}{2}}=\frac{7}{-\frac{11}{2}}$$

Dividing by -\frac{11}{2} undoes the multiplication by -\frac{11}{2}.

$$m^{2}+\frac{5}{-\frac{11}{2}}m=\frac{7}{-\frac{11}{2}}$$

Divide 5 by -\frac{11}{2} by multiplying 5 by the reciprocal of -\frac{11}{2}.

$$m^{2}-\frac{10}{11}m=\frac{7}{-\frac{11}{2}}$$

Divide 7 by -\frac{11}{2} by multiplying 7 by the reciprocal of -\frac{11}{2}.

$$m^{2}-\frac{10}{11}m=-\frac{14}{11}$$

Divide -\frac{10}{11}, the coefficient of the x term, by 2 to get -\frac{5}{11}. Then add the square of -\frac{5}{11} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$m^{2}-\frac{10}{11}m+\left(-\frac{5}{11}\right)^{2}=-\frac{14}{11}+\left(-\frac{5}{11}\right)^{2}$$

Square -\frac{5}{11} by squaring both the numerator and the denominator of the fraction.

$$m^{2}-\frac{10}{11}m+\frac{25}{121}=-\frac{14}{11}+\frac{25}{121}$$

Add -\frac{14}{11} to \frac{25}{121} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$m^{2}-\frac{10}{11}m+\frac{25}{121}=-\frac{129}{121}$$

Factor m^{2}-\frac{10}{11}m+\frac{25}{121}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(m-\frac{5}{11}\right)^{2}=-\frac{129}{121}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(m-\frac{5}{11}\right)^{2}}=\sqrt{-\frac{129}{121}}$$

Simplify.

$$m-\frac{5}{11}=\frac{\sqrt{129}i}{11} m-\frac{5}{11}=-\frac{\sqrt{129}i}{11}$$

Add \frac{5}{11} to both sides of the equation.

$$m=\frac{5+\sqrt{129}i}{11} m=\frac{-\sqrt{129}i+5}{11}$$