Solve 130t^2+180t+50=108

$$130 t ^ { 2 } + 180 t + 50 = 108$$

Solve for t

$$t=\frac{\sqrt{3910}}{65}-\frac{9}{13}\approx 0.269692197$$

$$t=-\frac{\sqrt{3910}}{65}-\frac{9}{13}\approx -1.654307582$$

Steps Using the Quadratic Formula

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

$$130t^{2}+180t+50=108$$

Subtract 108 from both sides of the equation.

$$130t^{2}+180t+50-108=108-108$$

Subtracting 108 from itself leaves 0.

$$130t^{2}+180t+50-108=0$$

Subtract 108 from 50.

$$130t^{2}+180t-58=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 130 for a, 180 for b, and -58 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$t=\frac{-180±\sqrt{180^{2}-4\times 130\left(-58\right)}}{2\times 130}$$

Square 180.

$$t=\frac{-180±\sqrt{32400-4\times 130\left(-58\right)}}{2\times 130}$$

Multiply -4 times 130.

$$t=\frac{-180±\sqrt{32400-520\left(-58\right)}}{2\times 130}$$

Multiply -520 times -58.

$$t=\frac{-180±\sqrt{32400+30160}}{2\times 130}$$

Add 32400 to 30160.

$$t=\frac{-180±\sqrt{62560}}{2\times 130}$$

Take the square root of 62560.

$$t=\frac{-180±4\sqrt{3910}}{2\times 130}$$

Multiply 2 times 130.

$$t=\frac{-180±4\sqrt{3910}}{260}$$

Now solve the equation t=\frac{-180±4\sqrt{3910}}{260} when ± is plus. Add -180 to 4\sqrt{3910}.

$$t=\frac{4\sqrt{3910}-180}{260}$$

Divide -180+4\sqrt{3910} by 260.

$$t=\frac{\sqrt{3910}}{65}-\frac{9}{13}$$

Now solve the equation t=\frac{-180±4\sqrt{3910}}{260} when ± is minus. Subtract 4\sqrt{3910} from -180.

$$t=\frac{-4\sqrt{3910}-180}{260}$$

Divide -180-4\sqrt{3910} by 260.

$$t=-\frac{\sqrt{3910}}{65}-\frac{9}{13}$$

The equation is now solved.

$$t=\frac{\sqrt{3910}}{65}-\frac{9}{13} t=-\frac{\sqrt{3910}}{65}-\frac{9}{13}$$

Steps for Completing the Square

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

$$130t^{2}+180t+50=108$$

Subtract 50 from both sides of the equation.

$$130t^{2}+180t+50-50=108-50$$

Subtracting 50 from itself leaves 0.

$$130t^{2}+180t=108-50$$

Subtract 50 from 108.

$$130t^{2}+180t=58$$

Divide both sides by 130.

$$\frac{130t^{2}+180t}{130}=\frac{58}{130}$$

Dividing by 130 undoes the multiplication by 130.

$$t^{2}+\frac{180}{130}t=\frac{58}{130}$$

Reduce the fraction \frac{180}{130} to lowest terms by extracting and canceling out 10.

$$t^{2}+\frac{18}{13}t=\frac{58}{130}$$

Reduce the fraction \frac{58}{130} to lowest terms by extracting and canceling out 2.

$$t^{2}+\frac{18}{13}t=\frac{29}{65}$$

Divide \frac{18}{13}, the coefficient of the x term, by 2 to get \frac{9}{13}. Then add the square of \frac{9}{13} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$t^{2}+\frac{18}{13}t+\left(\frac{9}{13}\right)^{2}=\frac{29}{65}+\left(\frac{9}{13}\right)^{2}$$

Square \frac{9}{13} by squaring both the numerator and the denominator of the fraction.

$$t^{2}+\frac{18}{13}t+\frac{81}{169}=\frac{29}{65}+\frac{81}{169}$$

Add \frac{29}{65} to \frac{81}{169} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$t^{2}+\frac{18}{13}t+\frac{81}{169}=\frac{782}{845}$$

Factor t^{2}+\frac{18}{13}t+\frac{81}{169}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(t+\frac{9}{13}\right)^{2}=\frac{782}{845}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(t+\frac{9}{13}\right)^{2}}=\sqrt{\frac{782}{845}}$$

Simplify.

$$t+\frac{9}{13}=\frac{\sqrt{3910}}{65} t+\frac{9}{13}=-\frac{\sqrt{3910}}{65}$$

Subtract \frac{9}{13} from both sides of the equation.

$$t=\frac{\sqrt{3910}}{65}-\frac{9}{13} t=-\frac{\sqrt{3910}}{65}-\frac{9}{13}$$