Solve 16t^2+9t-18*9=0

$$16 t ^ { 2 } + 9 t - 18 \times 9 = 0$$

Solve for t

$$t = \frac{9 \sqrt{129} - 9}{32} \approx 2.913135945$$

$$t=\frac{-9\sqrt{129}-9}{32}\approx -3.475635945$$

Steps Using the Quadratic Formula

Multiply 18 and 9 to get 162.

$$16t^{2}+9t-162=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 16 for a, 9 for b, and -162 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$t=\frac{-9±\sqrt{9^{2}-4\times 16\left(-162\right)}}{2\times 16}$$

Square 9.

$$t=\frac{-9±\sqrt{81-4\times 16\left(-162\right)}}{2\times 16}$$

Multiply -4 times 16.

$$t=\frac{-9±\sqrt{81-64\left(-162\right)}}{2\times 16}$$

Multiply -64 times -162.

$$t=\frac{-9±\sqrt{81+10368}}{2\times 16}$$

Add 81 to 10368.

$$t=\frac{-9±\sqrt{10449}}{2\times 16}$$

Take the square root of 10449.

$$t=\frac{-9±9\sqrt{129}}{2\times 16}$$

Multiply 2 times 16.

$$t=\frac{-9±9\sqrt{129}}{32}$$

Now solve the equation t=\frac{-9±9\sqrt{129}}{32} when ± is plus. Add -9 to 9\sqrt{129}.

$$t=\frac{9\sqrt{129}-9}{32}$$

Now solve the equation t=\frac{-9±9\sqrt{129}}{32} when ± is minus. Subtract 9\sqrt{129} from -9.

$$t=\frac{-9\sqrt{129}-9}{32}$$

The equation is now solved.

$$t=\frac{9\sqrt{129}-9}{32} t=\frac{-9\sqrt{129}-9}{32}$$

Steps for Completing the Square

Multiply 18 and 9 to get 162.

$$16t^{2}+9t-162=0$$

Add 162 to both sides. Anything plus zero gives itself.

$$16t^{2}+9t=162$$

Divide both sides by 16.

$$\frac{16t^{2}+9t}{16}=\frac{162}{16}$$

Dividing by 16 undoes the multiplication by 16.

$$t^{2}+\frac{9}{16}t=\frac{162}{16}$$

Reduce the fraction \frac{162}{16} to lowest terms by extracting and canceling out 2.

$$t^{2}+\frac{9}{16}t=\frac{81}{8}$$

Divide \frac{9}{16}, the coefficient of the x term, by 2 to get \frac{9}{32}. Then add the square of \frac{9}{32} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$t^{2}+\frac{9}{16}t+\left(\frac{9}{32}\right)^{2}=\frac{81}{8}+\left(\frac{9}{32}\right)^{2}$$

Square \frac{9}{32} by squaring both the numerator and the denominator of the fraction.

$$t^{2}+\frac{9}{16}t+\frac{81}{1024}=\frac{81}{8}+\frac{81}{1024}$$

Add \frac{81}{8} to \frac{81}{1024} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$t^{2}+\frac{9}{16}t+\frac{81}{1024}=\frac{10449}{1024}$$

Factor t^{2}+\frac{9}{16}t+\frac{81}{1024}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(t+\frac{9}{32}\right)^{2}=\frac{10449}{1024}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(t+\frac{9}{32}\right)^{2}}=\sqrt{\frac{10449}{1024}}$$

Simplify.

$$t+\frac{9}{32}=\frac{9\sqrt{129}}{32} t+\frac{9}{32}=-\frac{9\sqrt{129}}{32}$$

Subtract \frac{9}{32} from both sides of the equation.

$$t=\frac{9\sqrt{129}-9}{32} t=\frac{-9\sqrt{129}-9}{32}$$