Solve 4*9t^2+19*6t-2*9=0

$$4 \cdot 9 t ^ { 2 } + 19 \cdot 6 t - 2 \cdot 9 = 0$$

Solve for t

$$t=\frac{\sqrt{433}-19}{12}\approx 0.150721004$$

$$t=\frac{-\sqrt{433}-19}{12}\approx -3.317387671$$

Steps Using the Quadratic Formula

Do the multiplications.

$$36t^{2}+114t-2\times 9=0$$

Multiply 2 and 9 to get 18.

$$36t^{2}+114t-18=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 36 for a, 114 for b, and -18 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$t=\frac{-114±\sqrt{114^{2}-4\times 36\left(-18\right)}}{2\times 36}$$

Square 114.

$$t=\frac{-114±\sqrt{12996-4\times 36\left(-18\right)}}{2\times 36}$$

Multiply -4 times 36.

$$t=\frac{-114±\sqrt{12996-144\left(-18\right)}}{2\times 36}$$

Multiply -144 times -18.

$$t=\frac{-114±\sqrt{12996+2592}}{2\times 36}$$

Add 12996 to 2592.

$$t=\frac{-114±\sqrt{15588}}{2\times 36}$$

Take the square root of 15588.

$$t=\frac{-114±6\sqrt{433}}{2\times 36}$$

Multiply 2 times 36.

$$t=\frac{-114±6\sqrt{433}}{72}$$

Now solve the equation t=\frac{-114±6\sqrt{433}}{72} when ± is plus. Add -114 to 6\sqrt{433}.

$$t=\frac{6\sqrt{433}-114}{72}$$

Divide -114+6\sqrt{433} by 72.

$$t=\frac{\sqrt{433}-19}{12}$$

Now solve the equation t=\frac{-114±6\sqrt{433}}{72} when ± is minus. Subtract 6\sqrt{433} from -114.

$$t=\frac{-6\sqrt{433}-114}{72}$$

Divide -114-6\sqrt{433} by 72.

$$t=\frac{-\sqrt{433}-19}{12}$$

The equation is now solved.

$$t=\frac{\sqrt{433}-19}{12} t=\frac{-\sqrt{433}-19}{12}$$

Steps for Completing the Square

Do the multiplications.

$$36t^{2}+114t-2\times 9=0$$

Multiply 2 and 9 to get 18.

$$36t^{2}+114t-18=0$$

Add 18 to both sides. Anything plus zero gives itself.

$$36t^{2}+114t=18$$

Divide both sides by 36.

$$\frac{36t^{2}+114t}{36}=\frac{18}{36}$$

Dividing by 36 undoes the multiplication by 36.

$$t^{2}+\frac{114}{36}t=\frac{18}{36}$$

Reduce the fraction \frac{114}{36} to lowest terms by extracting and canceling out 6.

$$t^{2}+\frac{19}{6}t=\frac{18}{36}$$

Reduce the fraction \frac{18}{36} to lowest terms by extracting and canceling out 18.

$$t^{2}+\frac{19}{6}t=\frac{1}{2}$$

Divide \frac{19}{6}, the coefficient of the x term, by 2 to get \frac{19}{12}. Then add the square of \frac{19}{12} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$t^{2}+\frac{19}{6}t+\left(\frac{19}{12}\right)^{2}=\frac{1}{2}+\left(\frac{19}{12}\right)^{2}$$

Square \frac{19}{12} by squaring both the numerator and the denominator of the fraction.

$$t^{2}+\frac{19}{6}t+\frac{361}{144}=\frac{1}{2}+\frac{361}{144}$$

Add \frac{1}{2} to \frac{361}{144} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$t^{2}+\frac{19}{6}t+\frac{361}{144}=\frac{433}{144}$$

Factor t^{2}+\frac{19}{6}t+\frac{361}{144}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(t+\frac{19}{12}\right)^{2}=\frac{433}{144}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(t+\frac{19}{12}\right)^{2}}=\sqrt{\frac{433}{144}}$$

Simplify.

$$t+\frac{19}{12}=\frac{\sqrt{433}}{12} t+\frac{19}{12}=-\frac{\sqrt{433}}{12}$$

Subtract \frac{19}{12} from both sides of the equation.

$$t=\frac{\sqrt{433}-19}{12} t=\frac{-\sqrt{433}-19}{12}$$