Solve 5x^2=8x-16/5

$$5 x ^ { 2 } = 8 x - \frac { 16 } { 5 }$$

Solve for x

$$x=\frac{4}{5}=0.8$$

Steps Using the Quadratic Formula

Subtract 8x from both sides.

$$5x^{2}-8x=-\frac{16}{5}$$

Add \frac{16}{5} to both sides.

$$5x^{2}-8x+\frac{16}{5}=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 5 for a, -8 for b, and \frac{16}{5} for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-8\right)±\sqrt{\left(-8\right)^{2}-4\times 5\times \frac{16}{5}}}{2\times 5}$$

Square -8.

$$x=\frac{-\left(-8\right)±\sqrt{64-4\times 5\times \frac{16}{5}}}{2\times 5}$$

Multiply -4 times 5.

$$x=\frac{-\left(-8\right)±\sqrt{64-20\times \frac{16}{5}}}{2\times 5}$$

Multiply -20 times \frac{16}{5}.

$$x=\frac{-\left(-8\right)±\sqrt{64-64}}{2\times 5}$$

Add 64 to -64.

$$x=\frac{-\left(-8\right)±\sqrt{0}}{2\times 5}$$

Take the square root of 0.

$$x=-\frac{-8}{2\times 5}$$

The opposite of -8 is 8.

$$x=\frac{8}{2\times 5}$$

Multiply 2 times 5.

$$x=\frac{8}{10}$$

Reduce the fraction \frac{8}{10} to lowest terms by extracting and canceling out 2.

$$x=\frac{4}{5}$$

Steps for Completing the Square

Subtract 8x from both sides.

$$5x^{2}-8x=-\frac{16}{5}$$

Divide both sides by 5.

$$\frac{5x^{2}-8x}{5}=-\frac{\frac{16}{5}}{5}$$

Dividing by 5 undoes the multiplication by 5.

$$x^{2}-\frac{8}{5}x=-\frac{\frac{16}{5}}{5}$$

Divide -\frac{16}{5} by 5.

$$x^{2}-\frac{8}{5}x=-\frac{16}{25}$$

Divide -\frac{8}{5}, the coefficient of the x term, by 2 to get -\frac{4}{5}. Then add the square of -\frac{4}{5} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{8}{5}x+\left(-\frac{4}{5}\right)^{2}=-\frac{16}{25}+\left(-\frac{4}{5}\right)^{2}$$

Square -\frac{4}{5} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{8}{5}x+\frac{16}{25}=\frac{-16+16}{25}$$

Add -\frac{16}{25} to \frac{16}{25} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{8}{5}x+\frac{16}{25}=0$$

Factor x^{2}-\frac{8}{5}x+\frac{16}{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}.

$$\left(x-\frac{4}{5}\right)^{2}=0$$

Take the square root of both sides of the equation.

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

Simplify.

$$x-\frac{4}{5}=0 x-\frac{4}{5}=0$$

Add \frac{4}{5} to both sides of the equation.

$$x=\frac{4}{5} x=\frac{4}{5}$$

The equation is now solved. Solutions are the same.

$$x=\frac{4}{5}$$