Solve 98t^2=24*3

$$98 t ^ { 2 } = 24 \times 3$$

Solve for t

$$t=\frac{6}{7}\approx 0.857142857$$

$$t=-\frac{6}{7}\approx -0.857142857$$

Steps Using Difference of Squares

Multiply 24 and 3 to get 72.

$$98t^{2}=72$$

Subtract 72 from both sides.

$$98t^{2}-72=0$$

Divide both sides by 2.

$$49t^{2}-36=0$$

Consider 49t^{2}-36. Rewrite 49t^{2}-36 as \left(7t\right)^{2}-6^{2}. The difference of squares can be factored using the rule: a^{2}-b^{2}=\left(a-b\right)\left(a+b\right).

$$\left(7t-6\right)\left(7t+6\right)=0$$

To find equation solutions, solve 7t-6=0 and 7t+6=0.

$$t=\frac{6}{7} t=-\frac{6}{7}$$

Steps by Finding Square Root

Multiply 24 and 3 to get 72.

$$98t^{2}=72$$

Divide both sides by 98.

$$t^{2}=\frac{72}{98}$$

Reduce the fraction \frac{72}{98} to lowest terms by extracting and canceling out 2.

$$t^{2}=\frac{36}{49}$$

Take the square root of both sides of the equation.

$$t=\frac{6}{7} t=-\frac{6}{7}$$

Steps Using the Quadratic Formula

Multiply 24 and 3 to get 72.

$$98t^{2}=72$$

Subtract 72 from both sides.

$$98t^{2}-72=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 98 for a, 0 for b, and -72 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$t=\frac{0±\sqrt{0^{2}-4\times 98\left(-72\right)}}{2\times 98}$$

Square 0.

$$t=\frac{0±\sqrt{-4\times 98\left(-72\right)}}{2\times 98}$$

Multiply -4 times 98.

$$t=\frac{0±\sqrt{-392\left(-72\right)}}{2\times 98}$$

Multiply -392 times -72.

$$t=\frac{0±\sqrt{28224}}{2\times 98}$$

Take the square root of 28224.

$$t=\frac{0±168}{2\times 98}$$

Multiply 2 times 98.

$$t=\frac{0±168}{196}$$

Now solve the equation t=\frac{0±168}{196} when ± is plus. Reduce the fraction \frac{168}{196} to lowest terms by extracting and canceling out 28.

$$t=\frac{6}{7}$$

Now solve the equation t=\frac{0±168}{196} when ± is minus. Reduce the fraction \frac{-168}{196} to lowest terms by extracting and canceling out 28.

$$t=-\frac{6}{7}$$

The equation is now solved.

$$t=\frac{6}{7} t=-\frac{6}{7}$$