Solve 5d^2+9d+5=3d^2

$$5 d ^ { 2 } + 9 d + 5 = 3 d ^ { 2 }$$

Solve for d

$$d=\frac{\sqrt{41}-9}{4}\approx -0.649218941$$

$$d=\frac{-\sqrt{41}-9}{4}\approx -3.850781059$$

Steps Using the Quadratic Formula

Subtract 3d^{2} from both sides.

$$5d^{2}+9d+5-3d^{2}=0$$

Combine 5d^{2} and -3d^{2} to get 2d^{2}.

$$2d^{2}+9d+5=0$$

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

$$d=\frac{-9±\sqrt{9^{2}-4\times 2\times 5}}{2\times 2}$$

Square 9.

$$d=\frac{-9±\sqrt{81-4\times 2\times 5}}{2\times 2}$$

Multiply -4 times 2.

$$d=\frac{-9±\sqrt{81-8\times 5}}{2\times 2}$$

Multiply -8 times 5.

$$d=\frac{-9±\sqrt{81-40}}{2\times 2}$$

Add 81 to -40.

$$d=\frac{-9±\sqrt{41}}{2\times 2}$$

Multiply 2 times 2.

$$d=\frac{-9±\sqrt{41}}{4}$$

Now solve the equation d=\frac{-9±\sqrt{41}}{4} when ± is plus. Add -9 to \sqrt{41}.

$$d=\frac{\sqrt{41}-9}{4}$$

Now solve the equation d=\frac{-9±\sqrt{41}}{4} when ± is minus. Subtract \sqrt{41} from -9.

$$d=\frac{-\sqrt{41}-9}{4}$$

The equation is now solved.

$$d=\frac{\sqrt{41}-9}{4} d=\frac{-\sqrt{41}-9}{4}$$

Steps for Completing the Square

Subtract 3d^{2} from both sides.

$$5d^{2}+9d+5-3d^{2}=0$$

Combine 5d^{2} and -3d^{2} to get 2d^{2}.

$$2d^{2}+9d+5=0$$

Subtract 5 from both sides. Anything subtracted from zero gives its negation.

$$2d^{2}+9d=-5$$

Divide both sides by 2.

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

Dividing by 2 undoes the multiplication by 2.

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

Divide \frac{9}{2}, the coefficient of the x term, by 2 to get \frac{9}{4}. Then add the square of \frac{9}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$d^{2}+\frac{9}{2}d+\left(\frac{9}{4}\right)^{2}=-\frac{5}{2}+\left(\frac{9}{4}\right)^{2}$$

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

$$d^{2}+\frac{9}{2}d+\frac{81}{16}=-\frac{5}{2}+\frac{81}{16}$$

Add -\frac{5}{2} to \frac{81}{16} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$d^{2}+\frac{9}{2}d+\frac{81}{16}=\frac{41}{16}$$

Factor d^{2}+\frac{9}{2}d+\frac{81}{16}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(d+\frac{9}{4}\right)^{2}=\frac{41}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(d+\frac{9}{4}\right)^{2}}=\sqrt{\frac{41}{16}}$$

Simplify.

$$d+\frac{9}{4}=\frac{\sqrt{41}}{4} d+\frac{9}{4}=-\frac{\sqrt{41}}{4}$$

Subtract \frac{9}{4} from both sides of the equation.

$$d=\frac{\sqrt{41}-9}{4} d=\frac{-\sqrt{41}-9}{4}$$