Solve 5/4x^2-1/2x+025-65^2=0

$$\frac{ 5 }{ 4 } { x }^{ 2 } - \frac{ 1 }{ 2 } x+025- { 65 }^{ 2 } =0$$

Solve for x

$$x = \frac{3 \sqrt{9389} + 1}{5} \approx 58.338111424$$

$$x=\frac{1-3\sqrt{9389}}{5}\approx -57.938111424$$

Steps Using the Quadratic Formula

Multiply 0 and 25 to get 0.

$$\frac{5}{4}x^{2}-\frac{1}{2}x+0-65^{2}=0$$

Anything plus zero gives itself.

$$\frac{5}{4}x^{2}-\frac{1}{2}x-65^{2}=0$$

Calculate 65 to the power of 2 and get 4225.

$$\frac{5}{4}x^{2}-\frac{1}{2}x-4225=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute \frac{5}{4} for a, -\frac{1}{2} for b, and -4225 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-\frac{1}{2}\right)±\sqrt{\left(-\frac{1}{2}\right)^{2}-4\times \frac{5}{4}\left(-4225\right)}}{2\times \frac{5}{4}}$$

Square -\frac{1}{2} by squaring both the numerator and the denominator of the fraction.

$$x=\frac{-\left(-\frac{1}{2}\right)±\sqrt{\frac{1}{4}-4\times \frac{5}{4}\left(-4225\right)}}{2\times \frac{5}{4}}$$

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

$$x=\frac{-\left(-\frac{1}{2}\right)±\sqrt{\frac{1}{4}-5\left(-4225\right)}}{2\times \frac{5}{4}}$$

Multiply -5 times -4225.

$$x=\frac{-\left(-\frac{1}{2}\right)±\sqrt{\frac{1}{4}+21125}}{2\times \frac{5}{4}}$$

Add \frac{1}{4} to 21125.

$$x=\frac{-\left(-\frac{1}{2}\right)±\sqrt{\frac{84501}{4}}}{2\times \frac{5}{4}}$$

Take the square root of \frac{84501}{4}.

$$x=\frac{-\left(-\frac{1}{2}\right)±\frac{3\sqrt{9389}}{2}}{2\times \frac{5}{4}}$$

The opposite of -\frac{1}{2} is \frac{1}{2}.

$$x=\frac{\frac{1}{2}±\frac{3\sqrt{9389}}{2}}{2\times \frac{5}{4}}$$

Multiply 2 times \frac{5}{4}.

$$x=\frac{\frac{1}{2}±\frac{3\sqrt{9389}}{2}}{\frac{5}{2}}$$

Now solve the equation x=\frac{\frac{1}{2}±\frac{3\sqrt{9389}}{2}}{\frac{5}{2}} when ± is plus. Add \frac{1}{2} to \frac{3\sqrt{9389}}{2}.

$$x=\frac{3\sqrt{9389}+1}{2\times \frac{5}{2}}$$

Divide \frac{1+3\sqrt{9389}}{2} by \frac{5}{2} by multiplying \frac{1+3\sqrt{9389}}{2} by the reciprocal of \frac{5}{2}.

$$x=\frac{3\sqrt{9389}+1}{5}$$

Now solve the equation x=\frac{\frac{1}{2}±\frac{3\sqrt{9389}}{2}}{\frac{5}{2}} when ± is minus. Subtract \frac{3\sqrt{9389}}{2} from \frac{1}{2}.

$$x=\frac{1-3\sqrt{9389}}{2\times \frac{5}{2}}$$

Divide \frac{1-3\sqrt{9389}}{2} by \frac{5}{2} by multiplying \frac{1-3\sqrt{9389}}{2} by the reciprocal of \frac{5}{2}.

$$x=\frac{1-3\sqrt{9389}}{5}$$

The equation is now solved.

$$x=\frac{3\sqrt{9389}+1}{5} x=\frac{1-3\sqrt{9389}}{5}$$

Steps for Completing the Square

Multiply 0 and 25 to get 0.

$$\frac{5}{4}x^{2}-\frac{1}{2}x+0-65^{2}=0$$

Anything plus zero gives itself.

$$\frac{5}{4}x^{2}-\frac{1}{2}x-65^{2}=0$$

Calculate 65 to the power of 2 and get 4225.

$$\frac{5}{4}x^{2}-\frac{1}{2}x-4225=0$$

Add 4225 to both sides. Anything plus zero gives itself.

$$\frac{5}{4}x^{2}-\frac{1}{2}x=4225$$

Divide both sides of the equation by \frac{5}{4}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$\frac{\frac{5}{4}x^{2}-\frac{1}{2}x}{\frac{5}{4}}=\frac{4225}{\frac{5}{4}}$$

Dividing by \frac{5}{4} undoes the multiplication by \frac{5}{4}.

$$x^{2}+\left(-\frac{\frac{1}{2}}{\frac{5}{4}}\right)x=\frac{4225}{\frac{5}{4}}$$

Divide -\frac{1}{2} by \frac{5}{4} by multiplying -\frac{1}{2} by the reciprocal of \frac{5}{4}.

$$x^{2}-\frac{2}{5}x=\frac{4225}{\frac{5}{4}}$$

Divide 4225 by \frac{5}{4} by multiplying 4225 by the reciprocal of \frac{5}{4}.

$$x^{2}-\frac{2}{5}x=3380$$

Divide -\frac{2}{5}, the coefficient of the x term, by 2 to get -\frac{1}{5}. Then add the square of -\frac{1}{5} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{2}{5}x+\left(-\frac{1}{5}\right)^{2}=3380+\left(-\frac{1}{5}\right)^{2}$$

Square -\frac{1}{5} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{2}{5}x+\frac{1}{25}=3380+\frac{1}{25}$$

Add 3380 to \frac{1}{25}.

$$x^{2}-\frac{2}{5}x+\frac{1}{25}=\frac{84501}{25}$$

Factor x^{2}-\frac{2}{5}x+\frac{1}{25}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{1}{5}\right)^{2}=\frac{84501}{25}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{1}{5}\right)^{2}}=\sqrt{\frac{84501}{25}}$$

Simplify.

$$x-\frac{1}{5}=\frac{3\sqrt{9389}}{5} x-\frac{1}{5}=-\frac{3\sqrt{9389}}{5}$$

Add \frac{1}{5} to both sides of the equation.

$$x=\frac{3\sqrt{9389}+1}{5} x=\frac{1-3\sqrt{9389}}{5}$$