Solve =-5y^2/2+9y/2+3=0

$$= \frac { - 5 y ^ { 2 } } { 2 } + \frac { 9 y } { 2 } + 3 = 0$$

Solve for y

$$y = \frac{\sqrt{201} + 9}{10} \approx 2.317744688$$

$$y=\frac{9-\sqrt{201}}{10}\approx -0.517744688$$

Steps Using the Quadratic Formula

Multiply both sides of the equation by 2.

$$-5y^{2}+9y+6=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -5 for a, 9 for b, and 6 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$y=\frac{-9±\sqrt{9^{2}-4\left(-5\right)\times 6}}{2\left(-5\right)}$$

Square 9.

$$y=\frac{-9±\sqrt{81-4\left(-5\right)\times 6}}{2\left(-5\right)}$$

Multiply -4 times -5.

$$y=\frac{-9±\sqrt{81+20\times 6}}{2\left(-5\right)}$$

Multiply 20 times 6.

$$y=\frac{-9±\sqrt{81+120}}{2\left(-5\right)}$$

Add 81 to 120.

$$y=\frac{-9±\sqrt{201}}{2\left(-5\right)}$$

Multiply 2 times -5.

$$y=\frac{-9±\sqrt{201}}{-10}$$

Now solve the equation y=\frac{-9±\sqrt{201}}{-10} when ± is plus. Add -9 to \sqrt{201}.

$$y=\frac{\sqrt{201}-9}{-10}$$

Divide -9+\sqrt{201} by -10.

$$y=\frac{9-\sqrt{201}}{10}$$

Now solve the equation y=\frac{-9±\sqrt{201}}{-10} when ± is minus. Subtract \sqrt{201} from -9.

$$y=\frac{-\sqrt{201}-9}{-10}$$

Divide -9-\sqrt{201} by -10.

$$y=\frac{\sqrt{201}+9}{10}$$

The equation is now solved.

$$y=\frac{9-\sqrt{201}}{10} y=\frac{\sqrt{201}+9}{10}$$

Steps for Completing the Square

Multiply both sides of the equation by 2.

$$-5y^{2}+9y+6=0$$

Subtract 6 from both sides. Anything subtracted from zero gives its negation.

$$-5y^{2}+9y=-6$$

Divide both sides by -5.

$$\frac{-5y^{2}+9y}{-5}=-\frac{6}{-5}$$

Dividing by -5 undoes the multiplication by -5.

$$y^{2}+\frac{9}{-5}y=-\frac{6}{-5}$$

Divide 9 by -5.

$$y^{2}-\frac{9}{5}y=-\frac{6}{-5}$$

Divide -6 by -5.

$$y^{2}-\frac{9}{5}y=\frac{6}{5}$$

Divide -\frac{9}{5}, the coefficient of the x term, by 2 to get -\frac{9}{10}. Then add the square of -\frac{9}{10} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$y^{2}-\frac{9}{5}y+\left(-\frac{9}{10}\right)^{2}=\frac{6}{5}+\left(-\frac{9}{10}\right)^{2}$$

Square -\frac{9}{10} by squaring both the numerator and the denominator of the fraction.

$$y^{2}-\frac{9}{5}y+\frac{81}{100}=\frac{6}{5}+\frac{81}{100}$$

Add \frac{6}{5} to \frac{81}{100} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$y^{2}-\frac{9}{5}y+\frac{81}{100}=\frac{201}{100}$$

Factor y^{2}-\frac{9}{5}y+\frac{81}{100}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(y-\frac{9}{10}\right)^{2}=\frac{201}{100}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(y-\frac{9}{10}\right)^{2}}=\sqrt{\frac{201}{100}}$$

Simplify.

$$y-\frac{9}{10}=\frac{\sqrt{201}}{10} y-\frac{9}{10}=-\frac{\sqrt{201}}{10}$$

Add \frac{9}{10} to both sides of the equation.

$$y=\frac{\sqrt{201}+9}{10} y=\frac{9-\sqrt{201}}{10}$$