Solve 65=4x^2+5x-4

$$65=4 { x }^{ 2 } +5x-4$$

Solve for x

$$x = \frac{\sqrt{1129} - 5}{8} \approx 3.575074404$$

$$x=\frac{-\sqrt{1129}-5}{8}\approx -4.825074404$$

Steps Using the Quadratic Formula

Swap sides so that all variable terms are on the left hand side.

$$4x^{2}+5x-4=65$$

Subtract 65 from both sides.

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

Subtract 65 from -4 to get -69.

$$4x^{2}+5x-69=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 4 for a, 5 for b, and -69 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-5±\sqrt{5^{2}-4\times 4\left(-69\right)}}{2\times 4}$$

Square 5.

$$x=\frac{-5±\sqrt{25-4\times 4\left(-69\right)}}{2\times 4}$$

Multiply -4 times 4.

$$x=\frac{-5±\sqrt{25-16\left(-69\right)}}{2\times 4}$$

Multiply -16 times -69.

$$x=\frac{-5±\sqrt{25+1104}}{2\times 4}$$

Add 25 to 1104.

$$x=\frac{-5±\sqrt{1129}}{2\times 4}$$

Multiply 2 times 4.

$$x=\frac{-5±\sqrt{1129}}{8}$$

Now solve the equation x=\frac{-5±\sqrt{1129}}{8} when ± is plus. Add -5 to \sqrt{1129}.

$$x=\frac{\sqrt{1129}-5}{8}$$

Now solve the equation x=\frac{-5±\sqrt{1129}}{8} when ± is minus. Subtract \sqrt{1129} from -5.

$$x=\frac{-\sqrt{1129}-5}{8}$$

The equation is now solved.

$$x=\frac{\sqrt{1129}-5}{8} x=\frac{-\sqrt{1129}-5}{8}$$

Steps for Completing the Square

Swap sides so that all variable terms are on the left hand side.

$$4x^{2}+5x-4=65$$

Add 4 to both sides.

$$4x^{2}+5x=65+4$$

Add 65 and 4 to get 69.

$$4x^{2}+5x=69$$

Divide both sides by 4.

$$\frac{4x^{2}+5x}{4}=\frac{69}{4}$$

Dividing by 4 undoes the multiplication by 4.

$$x^{2}+\frac{5}{4}x=\frac{69}{4}$$

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

$$x^{2}+\frac{5}{4}x+\left(\frac{5}{8}\right)^{2}=\frac{69}{4}+\left(\frac{5}{8}\right)^{2}$$

Square \frac{5}{8} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{5}{4}x+\frac{25}{64}=\frac{69}{4}+\frac{25}{64}$$

Add \frac{69}{4} to \frac{25}{64} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{5}{4}x+\frac{25}{64}=\frac{1129}{64}$$

Factor x^{2}+\frac{5}{4}x+\frac{25}{64}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{5}{8}\right)^{2}=\frac{1129}{64}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{5}{8}\right)^{2}}=\sqrt{\frac{1129}{64}}$$

Simplify.

$$x+\frac{5}{8}=\frac{\sqrt{1129}}{8} x+\frac{5}{8}=-\frac{\sqrt{1129}}{8}$$

Subtract \frac{5}{8} from both sides of the equation.

$$x=\frac{\sqrt{1129}-5}{8} x=\frac{-\sqrt{1129}-5}{8}$$