Solve 8y(2y-5)=-25

$$8y(2y-5)=-25$$

Solve for y

$$y = \frac{5}{4} = 1\frac{1}{4} = 1.25$$

Steps Using the Quadratic Formula

Use the distributive property to multiply 8y by 2y-5.

$$16y^{2}-40y=-25$$

Add 25 to both sides.

$$16y^{2}-40y+25=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 16 for a, -40 for b, and 25 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$y=\frac{-\left(-40\right)±\sqrt{\left(-40\right)^{2}-4\times 16\times 25}}{2\times 16}$$

Square -40.

$$y=\frac{-\left(-40\right)±\sqrt{1600-4\times 16\times 25}}{2\times 16}$$

Multiply -4 times 16.

$$y=\frac{-\left(-40\right)±\sqrt{1600-64\times 25}}{2\times 16}$$

Multiply -64 times 25.

$$y=\frac{-\left(-40\right)±\sqrt{1600-1600}}{2\times 16}$$

Add 1600 to -1600.

$$y=\frac{-\left(-40\right)±\sqrt{0}}{2\times 16}$$

Take the square root of 0.

$$y=-\frac{-40}{2\times 16}$$

The opposite of -40 is 40.

$$y=\frac{40}{2\times 16}$$

Multiply 2 times 16.

$$y=\frac{40}{32}$$

Reduce the fraction \frac{40}{32} to lowest terms by extracting and canceling out 8.

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

Steps for Completing the Square

Use the distributive property to multiply 8y by 2y-5.

$$16y^{2}-40y=-25$$

Divide both sides by 16.

$$\frac{16y^{2}-40y}{16}=-\frac{25}{16}$$

Dividing by 16 undoes the multiplication by 16.

$$y^{2}+\left(-\frac{40}{16}\right)y=-\frac{25}{16}$$

Reduce the fraction \frac{-40}{16} to lowest terms by extracting and canceling out 8.

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

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

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

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

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

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

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

Factor y^{2}-\frac{5}{2}y+\frac{25}{16}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

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

Take the square root of both sides of the equation.

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

Simplify.

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

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

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

The equation is now solved. Solutions are the same.

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