Solve 7/10*5^2=10*0.5+7/10x^2

$$\frac{ 7 }{ 10 } \times { 5 }^{ 2 } = 10 \cdot 0.5+ \frac{ 7 }{ 10 } { x }^{ 2 }$$

Solve for x

$$x = \frac{5 \sqrt{35}}{7} \approx 4.225771274$$

$$x = -\frac{5 \sqrt{35}}{7} \approx -4.225771274$$

Steps by Finding Square Root

Calculate 5 to the power of 2 and get 25.

$$\frac{7}{10}\times 25=10\times 0.5+\frac{7}{10}x^{2}$$

Multiply \frac{7}{10} and 25 to get \frac{35}{2}.

$$\frac{35}{2}=10\times 0.5+\frac{7}{10}x^{2}$$

Multiply 10 and 0.5 to get 5.

$$\frac{35}{2}=5+\frac{7}{10}x^{2}$$

Swap sides so that all variable terms are on the left hand side.

$$5+\frac{7}{10}x^{2}=\frac{35}{2}$$

Subtract 5 from both sides.

$$\frac{7}{10}x^{2}=\frac{35}{2}-5$$

Subtract 5 from \frac{35}{2} to get \frac{25}{2}.

$$\frac{7}{10}x^{2}=\frac{25}{2}$$

Multiply both sides by \frac{10}{7}, the reciprocal of \frac{7}{10}.

$$x^{2}=\frac{25}{2}\times \frac{10}{7}$$

Multiply \frac{25}{2} and \frac{10}{7} to get \frac{125}{7}.

$$x^{2}=\frac{125}{7}$$

Take the square root of both sides of the equation.

$$x=\frac{5\sqrt{35}}{7} x=-\frac{5\sqrt{35}}{7}$$

Steps Using the Quadratic Formula

Calculate 5 to the power of 2 and get 25.

$$\frac{7}{10}\times 25=10\times 0.5+\frac{7}{10}x^{2}$$

Multiply \frac{7}{10} and 25 to get \frac{35}{2}.

$$\frac{35}{2}=10\times 0.5+\frac{7}{10}x^{2}$$

Multiply 10 and 0.5 to get 5.

$$\frac{35}{2}=5+\frac{7}{10}x^{2}$$

Swap sides so that all variable terms are on the left hand side.

$$5+\frac{7}{10}x^{2}=\frac{35}{2}$$

Subtract \frac{35}{2} from both sides.

$$5+\frac{7}{10}x^{2}-\frac{35}{2}=0$$

Subtract \frac{35}{2} from 5 to get -\frac{25}{2}.

$$-\frac{25}{2}+\frac{7}{10}x^{2}=0$$

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.

$$\frac{7}{10}x^{2}-\frac{25}{2}=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute \frac{7}{10} for a, 0 for b, and -\frac{25}{2} for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{0±\sqrt{0^{2}-4\times \frac{7}{10}\left(-\frac{25}{2}\right)}}{2\times \frac{7}{10}}$$

Square 0.

$$x=\frac{0±\sqrt{-4\times \frac{7}{10}\left(-\frac{25}{2}\right)}}{2\times \frac{7}{10}}$$

Multiply -4 times \frac{7}{10}.

$$x=\frac{0±\sqrt{-\frac{14}{5}\left(-\frac{25}{2}\right)}}{2\times \frac{7}{10}}$$

Multiply -\frac{14}{5} times -\frac{25}{2} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$x=\frac{0±\sqrt{35}}{2\times \frac{7}{10}}$$

Multiply 2 times \frac{7}{10}.

$$x=\frac{0±\sqrt{35}}{\frac{7}{5}}$$

Now solve the equation x=\frac{0±\sqrt{35}}{\frac{7}{5}} when ± is plus.

$$x=\frac{5\sqrt{35}}{7}$$

Now solve the equation x=\frac{0±\sqrt{35}}{\frac{7}{5}} when ± is minus.

$$x=-\frac{5\sqrt{35}}{7}$$

The equation is now solved.

$$x=\frac{5\sqrt{35}}{7} x=-\frac{5\sqrt{35}}{7}$$