Solve 15/4+0=-3/4b^2+3/2b+6

$$\frac { 15 } { 4 } + 0 = - \frac { 3 } { 4 } b ^ { 2 } + \frac { 3 } { 2 } b + 6$$

Solve for b

$$b=-1$$

$$b=3$$

Steps Using the Quadratic Formula

Add \frac{15}{4} and 0 to get \frac{15}{4}.

$$\frac{15}{4}=-\frac{3}{4}b^{2}+\frac{3}{2}b+6$$

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

$$-\frac{3}{4}b^{2}+\frac{3}{2}b+6=\frac{15}{4}$$

Subtract \frac{15}{4} from both sides.

$$-\frac{3}{4}b^{2}+\frac{3}{2}b+6-\frac{15}{4}=0$$

Subtract \frac{15}{4} from 6 to get \frac{9}{4}.

$$-\frac{3}{4}b^{2}+\frac{3}{2}b+\frac{9}{4}=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -\frac{3}{4} for a, \frac{3}{2} for b, and \frac{9}{4} for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

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

Square \frac{3}{2} by squaring both the numerator and the denominator of the fraction.

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

Multiply -4 times -\frac{3}{4}.

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

Multiply 3 times \frac{9}{4}.

$$b=\frac{-\frac{3}{2}±\sqrt{\frac{9+27}{4}}}{2\left(-\frac{3}{4}\right)}$$

Add \frac{9}{4} to \frac{27}{4} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

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

Take the square root of 9.

$$b=\frac{-\frac{3}{2}±3}{2\left(-\frac{3}{4}\right)}$$

Multiply 2 times -\frac{3}{4}.

$$b=\frac{-\frac{3}{2}±3}{-\frac{3}{2}}$$

Now solve the equation b=\frac{-\frac{3}{2}±3}{-\frac{3}{2}} when ± is plus. Add -\frac{3}{2} to 3.

$$b=\frac{\frac{3}{2}}{-\frac{3}{2}}$$

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

$$b=-1$$

Now solve the equation b=\frac{-\frac{3}{2}±3}{-\frac{3}{2}} when ± is minus. Subtract 3 from -\frac{3}{2}.

$$b=-\frac{\frac{9}{2}}{-\frac{3}{2}}$$

Divide -\frac{9}{2} by -\frac{3}{2} by multiplying -\frac{9}{2} by the reciprocal of -\frac{3}{2}.

$$b=3$$

The equation is now solved.

$$b=-1 b=3$$

Steps for Completing the Square

Add \frac{15}{4} and 0 to get \frac{15}{4}.

$$\frac{15}{4}=-\frac{3}{4}b^{2}+\frac{3}{2}b+6$$

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

$$-\frac{3}{4}b^{2}+\frac{3}{2}b+6=\frac{15}{4}$$

Subtract 6 from both sides.

$$-\frac{3}{4}b^{2}+\frac{3}{2}b=\frac{15}{4}-6$$

Subtract 6 from \frac{15}{4} to get -\frac{9}{4}.

$$-\frac{3}{4}b^{2}+\frac{3}{2}b=-\frac{9}{4}$$

Divide both sides of the equation by -\frac{3}{4}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$\frac{-\frac{3}{4}b^{2}+\frac{3}{2}b}{-\frac{3}{4}}=-\frac{\frac{9}{4}}{-\frac{3}{4}}$$

Dividing by -\frac{3}{4} undoes the multiplication by -\frac{3}{4}.

$$b^{2}+\frac{\frac{3}{2}}{-\frac{3}{4}}b=-\frac{\frac{9}{4}}{-\frac{3}{4}}$$

Divide \frac{3}{2} by -\frac{3}{4} by multiplying \frac{3}{2} by the reciprocal of -\frac{3}{4}.

$$b^{2}-2b=-\frac{\frac{9}{4}}{-\frac{3}{4}}$$

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

$$b^{2}-2b=3$$

Divide -2, the coefficient of the x term, by 2 to get -1. Then add the square of -1 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$b^{2}-2b+1=3+1$$

Add 3 to 1.

$$b^{2}-2b+1=4$$

Factor b^{2}-2b+1. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(b-1\right)^{2}=4$$

Take the square root of both sides of the equation.

$$\sqrt{\left(b-1\right)^{2}}=\sqrt{4}$$

Simplify.

$$b-1=2 b-1=-2$$

Add 1 to both sides of the equation.

$$b=3 b=-1$$