Solve 4y^2-28y=16

$$4 y ^ { 2 } - 28 y = 16$$

Solve for y

$$y = \frac{\sqrt{65} + 7}{2} \approx 7.531128874$$

$$y=\frac{7-\sqrt{65}}{2}\approx -0.531128874$$

Steps Using the Quadratic Formula

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

$$4y^{2}-28y=16$$

Subtract 16 from both sides of the equation.

$$4y^{2}-28y-16=16-16$$

Subtracting 16 from itself leaves 0.

$$4y^{2}-28y-16=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 4 for a, -28 for b, and -16 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$y=\frac{-\left(-28\right)±\sqrt{\left(-28\right)^{2}-4\times 4\left(-16\right)}}{2\times 4}$$

Square -28.

$$y=\frac{-\left(-28\right)±\sqrt{784-4\times 4\left(-16\right)}}{2\times 4}$$

Multiply -4 times 4.

$$y=\frac{-\left(-28\right)±\sqrt{784-16\left(-16\right)}}{2\times 4}$$

Multiply -16 times -16.

$$y=\frac{-\left(-28\right)±\sqrt{784+256}}{2\times 4}$$

Add 784 to 256.

$$y=\frac{-\left(-28\right)±\sqrt{1040}}{2\times 4}$$

Take the square root of 1040.

$$y=\frac{-\left(-28\right)±4\sqrt{65}}{2\times 4}$$

The opposite of -28 is 28.

$$y=\frac{28±4\sqrt{65}}{2\times 4}$$

Multiply 2 times 4.

$$y=\frac{28±4\sqrt{65}}{8}$$

Now solve the equation y=\frac{28±4\sqrt{65}}{8} when ± is plus. Add 28 to 4\sqrt{65}.

$$y=\frac{4\sqrt{65}+28}{8}$$

Divide 28+4\sqrt{65} by 8.

$$y=\frac{\sqrt{65}+7}{2}$$

Now solve the equation y=\frac{28±4\sqrt{65}}{8} when ± is minus. Subtract 4\sqrt{65} from 28.

$$y=\frac{28-4\sqrt{65}}{8}$$

Divide 28-4\sqrt{65} by 8.

$$y=\frac{7-\sqrt{65}}{2}$$

The equation is now solved.

$$y=\frac{\sqrt{65}+7}{2} y=\frac{7-\sqrt{65}}{2}$$

Steps for Completing the Square

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

$$4y^{2}-28y=16$$

Divide both sides by 4.

$$\frac{4y^{2}-28y}{4}=\frac{16}{4}$$

Dividing by 4 undoes the multiplication by 4.

$$y^{2}+\left(-\frac{28}{4}\right)y=\frac{16}{4}$$

Divide -28 by 4.

$$y^{2}-7y=\frac{16}{4}$$

Divide 16 by 4.

$$y^{2}-7y=4$$

Divide -7, the coefficient of the x term, by 2 to get -\frac{7}{2}. Then add the square of -\frac{7}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$y^{2}-7y+\left(-\frac{7}{2}\right)^{2}=4+\left(-\frac{7}{2}\right)^{2}$$

Square -\frac{7}{2} by squaring both the numerator and the denominator of the fraction.

$$y^{2}-7y+\frac{49}{4}=4+\frac{49}{4}$$

Add 4 to \frac{49}{4}.

$$y^{2}-7y+\frac{49}{4}=\frac{65}{4}$$

Factor y^{2}-7y+\frac{49}{4}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(y-\frac{7}{2}\right)^{2}=\frac{65}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(y-\frac{7}{2}\right)^{2}}=\sqrt{\frac{65}{4}}$$

Simplify.

$$y-\frac{7}{2}=\frac{\sqrt{65}}{2} y-\frac{7}{2}=-\frac{\sqrt{65}}{2}$$

Add \frac{7}{2} to both sides of the equation.

$$y=\frac{\sqrt{65}+7}{2} y=\frac{7-\sqrt{65}}{2}$$