Solve 4.9t^2-1.38=0

$$4.9 t ^ { 2 } - 1.38 = 0$$

Solve for t

$$t=\frac{\sqrt{345}}{35}\approx 0.530690732$$

$$t=-\frac{\sqrt{345}}{35}\approx -0.530690732$$

Steps by Finding Square Root

Add 1.38 to both sides. Anything plus zero gives itself.

$$4.9t^{2}=1.38$$

Divide both sides by 4.9.

$$t^{2}=\frac{1.38}{4.9}$$

Expand \frac{1.38}{4.9} by multiplying both numerator and the denominator by 100.

$$t^{2}=\frac{138}{490}$$

Reduce the fraction \frac{138}{490} to lowest terms by extracting and canceling out 2.

$$t^{2}=\frac{69}{245}$$

Take the square root of both sides of the equation.

$$t=\frac{\sqrt{345}}{35} t=-\frac{\sqrt{345}}{35}$$

Steps Using the Quadratic Formula

Quadratic equations like this one, with an x^{2} term but no x term, can still be solved using the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}, once they are put in standard form: ax^{2}+bx+c=0.

$$4.9t^{2}-1.38=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 4.9 for a, 0 for b, and -1.38 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$t=\frac{0±\sqrt{0^{2}-4\times 4.9\left(-1.38\right)}}{2\times 4.9}$$

Square 0.

$$t=\frac{0±\sqrt{-4\times 4.9\left(-1.38\right)}}{2\times 4.9}$$

Multiply -4 times 4.9.

$$t=\frac{0±\sqrt{-19.6\left(-1.38\right)}}{2\times 4.9}$$

Multiply -19.6 times -1.38 by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$t=\frac{0±\sqrt{27.048}}{2\times 4.9}$$

Take the square root of 27.048.

$$t=\frac{0±\frac{7\sqrt{345}}{25}}{2\times 4.9}$$

Multiply 2 times 4.9.

$$t=\frac{0±\frac{7\sqrt{345}}{25}}{9.8}$$

Now solve the equation t=\frac{0±\frac{7\sqrt{345}}{25}}{9.8} when ± is plus.

$$t=\frac{\sqrt{345}}{35}$$

Now solve the equation t=\frac{0±\frac{7\sqrt{345}}{25}}{9.8} when ± is minus.

$$t=-\frac{\sqrt{345}}{35}$$

The equation is now solved.

$$t=\frac{\sqrt{345}}{35} t=-\frac{\sqrt{345}}{35}$$