Solve 41x^2-984x+5904=0

$$41 { x }^{ 2 } -984x+5904=0$$

Solve for x

$$x=12$$

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.

$$41x^{2}-984x+5904=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 41 for a, -984 for b, and 5904 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-984\right)±\sqrt{\left(-984\right)^{2}-4\times 41\times 5904}}{2\times 41}$$

Square -984.

$$x=\frac{-\left(-984\right)±\sqrt{968256-4\times 41\times 5904}}{2\times 41}$$

Multiply -4 times 41.

$$x=\frac{-\left(-984\right)±\sqrt{968256-164\times 5904}}{2\times 41}$$

Multiply -164 times 5904.

$$x=\frac{-\left(-984\right)±\sqrt{968256-968256}}{2\times 41}$$

Add 968256 to -968256.

$$x=\frac{-\left(-984\right)±\sqrt{0}}{2\times 41}$$

Take the square root of 0.

$$x=-\frac{-984}{2\times 41}$$

The opposite of -984 is 984.

$$x=\frac{984}{2\times 41}$$

Multiply 2 times 41.

$$x=\frac{984}{82}$$

Divide 984 by 82.

$$x=12$$

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.

$$41x^{2}-984x+5904=0$$

Subtract 5904 from both sides of the equation.

$$41x^{2}-984x+5904-5904=-5904$$

Subtracting 5904 from itself leaves 0.

$$41x^{2}-984x=-5904$$

Divide both sides by 41.

$$\frac{41x^{2}-984x}{41}=-\frac{5904}{41}$$

Dividing by 41 undoes the multiplication by 41.

$$x^{2}+\left(-\frac{984}{41}\right)x=-\frac{5904}{41}$$

Divide -984 by 41.

$$x^{2}-24x=-\frac{5904}{41}$$

Divide -5904 by 41.

$$x^{2}-24x=-144$$

Divide -24, the coefficient of the x term, by 2 to get -12. Then add the square of -12 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-24x+\left(-12\right)^{2}=-144+\left(-12\right)^{2}$$

Square -12.

$$x^{2}-24x+144=-144+144$$

Add -144 to 144.

$$x^{2}-24x+144=0$$

Factor x^{2}-24x+144. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-12\right)^{2}=0$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-12\right)^{2}}=\sqrt{0}$$

Simplify.

$$x-12=0 x-12=0$$

Add 12 to both sides of the equation.

$$x=12 x=12$$

The equation is now solved. Solutions are the same.

$$x=12$$