Solve x^2div(2*9.8)=9.3

$${ x }^{ 2 } \div (2 \times 9.8)=9.3$$

Solve for x

$$x = \frac{7 \sqrt{93}}{5} \approx 13.501111065$$

$$x = -\frac{7 \sqrt{93}}{5} \approx -13.501111065$$

Steps by Finding Square Root

Multiply 2 and 9.8 to get 19.6.

$$\frac{x^{2}}{19.6}=9.3$$

Multiply both sides by 19.6.

$$x^{2}=9.3\times 19.6$$

Multiply 9.3 and 19.6 to get 182.28.

$$x^{2}=182.28$$

Take the square root of both sides of the equation.

$$x=\frac{7\sqrt{93}}{5} x=-\frac{7\sqrt{93}}{5}$$

Steps Using the Quadratic Formula

Multiply 2 and 9.8 to get 19.6.

$$\frac{x^{2}}{19.6}=9.3$$

Subtract 9.3 from both sides.

$$\frac{x^{2}}{19.6}-9.3=0$$

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.

$$\frac{5}{98}x^{2}-9.3=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute \frac{5}{98} for a, 0 for b, and -9.3 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{0±\sqrt{0^{2}-4\times \frac{5}{98}\left(-9.3\right)}}{2\times \frac{5}{98}}$$

Square 0.

$$x=\frac{0±\sqrt{-4\times \frac{5}{98}\left(-9.3\right)}}{2\times \frac{5}{98}}$$

Multiply -4 times \frac{5}{98}.

$$x=\frac{0±\sqrt{-\frac{10}{49}\left(-9.3\right)}}{2\times \frac{5}{98}}$$

Multiply -\frac{10}{49} times -9.3 by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$x=\frac{0±\sqrt{\frac{93}{49}}}{2\times \frac{5}{98}}$$

Take the square root of \frac{93}{49}.

$$x=\frac{0±\frac{\sqrt{93}}{7}}{2\times \frac{5}{98}}$$

Multiply 2 times \frac{5}{98}.

$$x=\frac{0±\frac{\sqrt{93}}{7}}{\frac{5}{49}}$$

Now solve the equation x=\frac{0±\frac{\sqrt{93}}{7}}{\frac{5}{49}} when ± is plus.

$$x=\frac{7\sqrt{93}}{5}$$

Now solve the equation x=\frac{0±\frac{\sqrt{93}}{7}}{\frac{5}{49}} when ± is minus.

$$x=-\frac{7\sqrt{93}}{5}$$

The equation is now solved.

$$x=\frac{7\sqrt{93}}{5} x=-\frac{7\sqrt{93}}{5}$$