Solve -35t-1/2*98t^2=-14

$$- 35 t - \frac { 1 } { 2 } \times 98 t ^ { 2 } = - 14$$

Solve for t

$$t=-1$$

$$t=\frac{2}{7}\approx 0.285714286$$

Steps Using Factoring By Grouping

Multiply \frac{1}{2} and 98 to get 49.

$$-35t-49t^{2}=-14$$

Add 14 to both sides.

$$-35t-49t^{2}+14=0$$

Divide both sides by 7.

$$-5t-7t^{2}+2=0$$

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$-7t^{2}-5t+2=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -7t^{2}+at+bt+2. To find a and b, set up a system to be solved.

$$a+b=-5 ab=-7\times 2=-14$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -14.

$$1,-14 2,-7$$

Calculate the sum for each pair.

$$1-14=-13 2-7=-5$$

The solution is the pair that gives sum -5.

$$a=2 b=-7$$

Rewrite -7t^{2}-5t+2 as \left(-7t^{2}+2t\right)+\left(-7t+2\right).

$$\left(-7t^{2}+2t\right)+\left(-7t+2\right)$$

Factor out -t in the first and -1 in the second group.

$$-t\left(7t-2\right)-\left(7t-2\right)$$

Factor out common term 7t-2 by using distributive property.

$$\left(7t-2\right)\left(-t-1\right)$$

To find equation solutions, solve 7t-2=0 and -t-1=0.

$$t=\frac{2}{7} t=-1$$

Steps Using the Quadratic Formula

Multiply \frac{1}{2} and 98 to get 49.

$$-35t-49t^{2}=-14$$

Add 14 to both sides.

$$-35t-49t^{2}+14=0$$

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.

$$-49t^{2}-35t+14=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -49 for a, -35 for b, and 14 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$t=\frac{-\left(-35\right)±\sqrt{\left(-35\right)^{2}-4\left(-49\right)\times 14}}{2\left(-49\right)}$$

Square -35.

$$t=\frac{-\left(-35\right)±\sqrt{1225-4\left(-49\right)\times 14}}{2\left(-49\right)}$$

Multiply -4 times -49.

$$t=\frac{-\left(-35\right)±\sqrt{1225+196\times 14}}{2\left(-49\right)}$$

Multiply 196 times 14.

$$t=\frac{-\left(-35\right)±\sqrt{1225+2744}}{2\left(-49\right)}$$

Add 1225 to 2744.

$$t=\frac{-\left(-35\right)±\sqrt{3969}}{2\left(-49\right)}$$

Take the square root of 3969.

$$t=\frac{-\left(-35\right)±63}{2\left(-49\right)}$$

The opposite of -35 is 35.

$$t=\frac{35±63}{2\left(-49\right)}$$

Multiply 2 times -49.

$$t=\frac{35±63}{-98}$$

Now solve the equation t=\frac{35±63}{-98} when ± is plus. Add 35 to 63.

$$t=\frac{98}{-98}$$

Divide 98 by -98.

$$t=-1$$

Now solve the equation t=\frac{35±63}{-98} when ± is minus. Subtract 63 from 35.

$$t=-\frac{28}{-98}$$

Reduce the fraction \frac{-28}{-98} to lowest terms by extracting and canceling out 14.

$$t=\frac{2}{7}$$

The equation is now solved.

$$t=-1 t=\frac{2}{7}$$

Steps for Completing the Square

Multiply \frac{1}{2} and 98 to get 49.

$$-35t-49t^{2}=-14$$

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.

$$-49t^{2}-35t=-14$$

Divide both sides by -49.

$$\frac{-49t^{2}-35t}{-49}=-\frac{14}{-49}$$

Dividing by -49 undoes the multiplication by -49.

$$t^{2}+\left(-\frac{35}{-49}\right)t=-\frac{14}{-49}$$

Reduce the fraction \frac{-35}{-49} to lowest terms by extracting and canceling out 7.

$$t^{2}+\frac{5}{7}t=-\frac{14}{-49}$$

Reduce the fraction \frac{-14}{-49} to lowest terms by extracting and canceling out 7.

$$t^{2}+\frac{5}{7}t=\frac{2}{7}$$

Divide \frac{5}{7}, the coefficient of the x term, by 2 to get \frac{5}{14}. Then add the square of \frac{5}{14} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$t^{2}+\frac{5}{7}t+\left(\frac{5}{14}\right)^{2}=\frac{2}{7}+\left(\frac{5}{14}\right)^{2}$$

Square \frac{5}{14} by squaring both the numerator and the denominator of the fraction.

$$t^{2}+\frac{5}{7}t+\frac{25}{196}=\frac{2}{7}+\frac{25}{196}$$

Add \frac{2}{7} to \frac{25}{196} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$t^{2}+\frac{5}{7}t+\frac{25}{196}=\frac{81}{196}$$

Factor t^{2}+\frac{5}{7}t+\frac{25}{196}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(t+\frac{5}{14}\right)^{2}=\frac{81}{196}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(t+\frac{5}{14}\right)^{2}}=\sqrt{\frac{81}{196}}$$

Simplify.

$$t+\frac{5}{14}=\frac{9}{14} t+\frac{5}{14}=-\frac{9}{14}$$

Subtract \frac{5}{14} from both sides of the equation.

$$t=\frac{2}{7} t=-1$$