Solve 1:4x^2-9x=28

$$1 : 4 x ^ { 2 } - 9 x = 28$$

Solve for x

$$x=2\sqrt{109}+18\approx 38.880613018$$

$$x=18-2\sqrt{109}\approx -2.880613018$$

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.

$$\frac{1}{4}x^{2}-9x=28$$

Subtract 28 from both sides of the equation.

$$\frac{1}{4}x^{2}-9x-28=28-28$$

Subtracting 28 from itself leaves 0.

$$\frac{1}{4}x^{2}-9x-28=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute \frac{1}{4} for a, -9 for b, and -28 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-9\right)±\sqrt{\left(-9\right)^{2}-4\times \frac{1}{4}\left(-28\right)}}{2\times \frac{1}{4}}$$

Square -9.

$$x=\frac{-\left(-9\right)±\sqrt{81-4\times \frac{1}{4}\left(-28\right)}}{2\times \frac{1}{4}}$$

Multiply -4 times \frac{1}{4}.

$$x=\frac{-\left(-9\right)±\sqrt{81-\left(-28\right)}}{2\times \frac{1}{4}}$$

Multiply -1 times -28.

$$x=\frac{-\left(-9\right)±\sqrt{81+28}}{2\times \frac{1}{4}}$$

Add 81 to 28.

$$x=\frac{-\left(-9\right)±\sqrt{109}}{2\times \frac{1}{4}}$$

The opposite of -9 is 9.

$$x=\frac{9±\sqrt{109}}{2\times \frac{1}{4}}$$

Multiply 2 times \frac{1}{4}.

$$x=\frac{9±\sqrt{109}}{\frac{1}{2}}$$

Now solve the equation x=\frac{9±\sqrt{109}}{\frac{1}{2}} when ± is plus. Add 9 to \sqrt{109}.

$$x=\frac{\sqrt{109}+9}{\frac{1}{2}}$$

Divide 9+\sqrt{109} by \frac{1}{2} by multiplying 9+\sqrt{109} by the reciprocal of \frac{1}{2}.

$$x=2\sqrt{109}+18$$

Now solve the equation x=\frac{9±\sqrt{109}}{\frac{1}{2}} when ± is minus. Subtract \sqrt{109} from 9.

$$x=\frac{9-\sqrt{109}}{\frac{1}{2}}$$

Divide 9-\sqrt{109} by \frac{1}{2} by multiplying 9-\sqrt{109} by the reciprocal of \frac{1}{2}.

$$x=18-2\sqrt{109}$$

The equation is now solved.

$$x=2\sqrt{109}+18 x=18-2\sqrt{109}$$

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.

$$\frac{1}{4}x^{2}-9x=28$$

Multiply both sides by 4.

$$\frac{\frac{1}{4}x^{2}-9x}{\frac{1}{4}}=\frac{28}{\frac{1}{4}}$$

Dividing by \frac{1}{4} undoes the multiplication by \frac{1}{4}.

$$x^{2}+\left(-\frac{9}{\frac{1}{4}}\right)x=\frac{28}{\frac{1}{4}}$$

Divide -9 by \frac{1}{4} by multiplying -9 by the reciprocal of \frac{1}{4}.

$$x^{2}-36x=\frac{28}{\frac{1}{4}}$$

Divide 28 by \frac{1}{4} by multiplying 28 by the reciprocal of \frac{1}{4}.

$$x^{2}-36x=112$$

Divide -36, the coefficient of the x term, by 2 to get -18. Then add the square of -18 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-36x+\left(-18\right)^{2}=112+\left(-18\right)^{2}$$

Square -18.

$$x^{2}-36x+324=112+324$$

Add 112 to 324.

$$x^{2}-36x+324=436$$

Factor x^{2}-36x+324. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-18\right)^{2}=436$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-18\right)^{2}}=\sqrt{436}$$

Simplify.

$$x-18=2\sqrt{109} x-18=-2\sqrt{109}$$

Add 18 to both sides of the equation.

$$x=2\sqrt{109}+18 x=18-2\sqrt{109}$$