Solve x^2-4x^2+40x+64=0

$$x ^ { 2 } - 4 x ^ { 2 } + 40 x + 64 = 0$$

Solve for x

$$x = \frac{4 \sqrt{37} + 20}{3} \approx 14.777016707$$

$$x=\frac{20-4\sqrt{37}}{3}\approx -1.443683374$$

Steps Using the Quadratic Formula

Combine x^{2} and -4x^{2} to get -3x^{2}.

$$-3x^{2}+40x+64=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -3 for a, 40 for b, and 64 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-40±\sqrt{40^{2}-4\left(-3\right)\times 64}}{2\left(-3\right)}$$

Square 40.

$$x=\frac{-40±\sqrt{1600-4\left(-3\right)\times 64}}{2\left(-3\right)}$$

Multiply -4 times -3.

$$x=\frac{-40±\sqrt{1600+12\times 64}}{2\left(-3\right)}$$

Multiply 12 times 64.

$$x=\frac{-40±\sqrt{1600+768}}{2\left(-3\right)}$$

Add 1600 to 768.

$$x=\frac{-40±\sqrt{2368}}{2\left(-3\right)}$$

Take the square root of 2368.

$$x=\frac{-40±8\sqrt{37}}{2\left(-3\right)}$$

Multiply 2 times -3.

$$x=\frac{-40±8\sqrt{37}}{-6}$$

Now solve the equation x=\frac{-40±8\sqrt{37}}{-6} when ± is plus. Add -40 to 8\sqrt{37}.

$$x=\frac{8\sqrt{37}-40}{-6}$$

Divide -40+8\sqrt{37} by -6.

$$x=\frac{20-4\sqrt{37}}{3}$$

Now solve the equation x=\frac{-40±8\sqrt{37}}{-6} when ± is minus. Subtract 8\sqrt{37} from -40.

$$x=\frac{-8\sqrt{37}-40}{-6}$$

Divide -40-8\sqrt{37} by -6.

$$x=\frac{4\sqrt{37}+20}{3}$$

The equation is now solved.

$$x=\frac{20-4\sqrt{37}}{3} x=\frac{4\sqrt{37}+20}{3}$$

Steps for Completing the Square

Combine x^{2} and -4x^{2} to get -3x^{2}.

$$-3x^{2}+40x+64=0$$

Subtract 64 from both sides. Anything subtracted from zero gives its negation.

$$-3x^{2}+40x=-64$$

Divide both sides by -3.

$$\frac{-3x^{2}+40x}{-3}=-\frac{64}{-3}$$

Dividing by -3 undoes the multiplication by -3.

$$x^{2}+\frac{40}{-3}x=-\frac{64}{-3}$$

Divide 40 by -3.

$$x^{2}-\frac{40}{3}x=-\frac{64}{-3}$$

Divide -64 by -3.

$$x^{2}-\frac{40}{3}x=\frac{64}{3}$$

Divide -\frac{40}{3}, the coefficient of the x term, by 2 to get -\frac{20}{3}. Then add the square of -\frac{20}{3} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{40}{3}x+\left(-\frac{20}{3}\right)^{2}=\frac{64}{3}+\left(-\frac{20}{3}\right)^{2}$$

Square -\frac{20}{3} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{40}{3}x+\frac{400}{9}=\frac{64}{3}+\frac{400}{9}$$

Add \frac{64}{3} to \frac{400}{9} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{40}{3}x+\frac{400}{9}=\frac{592}{9}$$

Factor x^{2}-\frac{40}{3}x+\frac{400}{9}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{20}{3}\right)^{2}=\frac{592}{9}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{20}{3}\right)^{2}}=\sqrt{\frac{592}{9}}$$

Simplify.

$$x-\frac{20}{3}=\frac{4\sqrt{37}}{3} x-\frac{20}{3}=-\frac{4\sqrt{37}}{3}$$

Add \frac{20}{3} to both sides of the equation.

$$x=\frac{4\sqrt{37}+20}{3} x=\frac{20-4\sqrt{37}}{3}$$