Solve -51t+49t^2=105

$$-51t+49 { t }^{ 2 } = 105$$

Solve for t

$$t = \frac{\sqrt{23181} + 51}{98} \approx 2.074011008$$

$$t=\frac{51-\sqrt{23181}}{98}\approx -1.033194681$$

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.

$$49t^{2}-51t=105$$

Subtract 105 from both sides of the equation.

$$49t^{2}-51t-105=105-105$$

Subtracting 105 from itself leaves 0.

$$49t^{2}-51t-105=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 49 for a, -51 for b, and -105 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$t=\frac{-\left(-51\right)±\sqrt{\left(-51\right)^{2}-4\times 49\left(-105\right)}}{2\times 49}$$

Square -51.

$$t=\frac{-\left(-51\right)±\sqrt{2601-4\times 49\left(-105\right)}}{2\times 49}$$

Multiply -4 times 49.

$$t=\frac{-\left(-51\right)±\sqrt{2601-196\left(-105\right)}}{2\times 49}$$

Multiply -196 times -105.

$$t=\frac{-\left(-51\right)±\sqrt{2601+20580}}{2\times 49}$$

Add 2601 to 20580.

$$t=\frac{-\left(-51\right)±\sqrt{23181}}{2\times 49}$$

The opposite of -51 is 51.

$$t=\frac{51±\sqrt{23181}}{2\times 49}$$

Multiply 2 times 49.

$$t=\frac{51±\sqrt{23181}}{98}$$

Now solve the equation t=\frac{51±\sqrt{23181}}{98} when ± is plus. Add 51 to \sqrt{23181}.

$$t=\frac{\sqrt{23181}+51}{98}$$

Now solve the equation t=\frac{51±\sqrt{23181}}{98} when ± is minus. Subtract \sqrt{23181} from 51.

$$t=\frac{51-\sqrt{23181}}{98}$$

The equation is now solved.

$$t=\frac{\sqrt{23181}+51}{98} t=\frac{51-\sqrt{23181}}{98}$$

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.

$$49t^{2}-51t=105$$

Divide both sides by 49.

$$\frac{49t^{2}-51t}{49}=\frac{105}{49}$$

Dividing by 49 undoes the multiplication by 49.

$$t^{2}-\frac{51}{49}t=\frac{105}{49}$$

Reduce the fraction \frac{105}{49} to lowest terms by extracting and canceling out 7.

$$t^{2}-\frac{51}{49}t=\frac{15}{7}$$

Divide -\frac{51}{49}, the coefficient of the x term, by 2 to get -\frac{51}{98}. Then add the square of -\frac{51}{98} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$t^{2}-\frac{51}{49}t+\left(-\frac{51}{98}\right)^{2}=\frac{15}{7}+\left(-\frac{51}{98}\right)^{2}$$

Square -\frac{51}{98} by squaring both the numerator and the denominator of the fraction.

$$t^{2}-\frac{51}{49}t+\frac{2601}{9604}=\frac{15}{7}+\frac{2601}{9604}$$

Add \frac{15}{7} to \frac{2601}{9604} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$t^{2}-\frac{51}{49}t+\frac{2601}{9604}=\frac{23181}{9604}$$

Factor t^{2}-\frac{51}{49}t+\frac{2601}{9604}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(t-\frac{51}{98}\right)^{2}=\frac{23181}{9604}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(t-\frac{51}{98}\right)^{2}}=\sqrt{\frac{23181}{9604}}$$

Simplify.

$$t-\frac{51}{98}=\frac{\sqrt{23181}}{98} t-\frac{51}{98}=-\frac{\sqrt{23181}}{98}$$

Add \frac{51}{98} to both sides of the equation.

$$t=\frac{\sqrt{23181}+51}{98} t=\frac{51-\sqrt{23181}}{98}$$