x^2+54x-5=500-এর সমাধান করুন

$${ x }^{ 2 } +54x-5=500$$

Solve for x (complex solution)

$$x=\sqrt{1234}-27\approx 8.128336141$$

$$x=-\left(\sqrt{1234}+27\right)\approx -62.128336141$$

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.

$$x^{2}+54x-5=500$$

Subtract 500 from both sides of the equation.

$$x^{2}+54x-5-500=500-500$$

Subtracting 500 from itself leaves 0.

$$x^{2}+54x-5-500=0$$

Subtract 500 from -5.

$$x^{2}+54x-505=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 54 for b, and -505 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-54±\sqrt{54^{2}-4\left(-505\right)}}{2}$$

Square 54.

$$x=\frac{-54±\sqrt{2916-4\left(-505\right)}}{2}$$

Multiply -4 times -505.

$$x=\frac{-54±\sqrt{2916+2020}}{2}$$

Add 2916 to 2020.

$$x=\frac{-54±\sqrt{4936}}{2}$$

Take the square root of 4936.

$$x=\frac{-54±2\sqrt{1234}}{2}$$

Now solve the equation x=\frac{-54±2\sqrt{1234}}{2} when ± is plus. Add -54 to 2\sqrt{1234}.

$$x=\frac{2\sqrt{1234}-54}{2}$$

Divide -54+2\sqrt{1234} by 2.

$$x=\sqrt{1234}-27$$

Now solve the equation x=\frac{-54±2\sqrt{1234}}{2} when ± is minus. Subtract 2\sqrt{1234} from -54.

$$x=\frac{-2\sqrt{1234}-54}{2}$$

Divide -54-2\sqrt{1234} by 2.

$$x=-\sqrt{1234}-27$$

The equation is now solved.

$$x=\sqrt{1234}-27 x=-\sqrt{1234}-27$$

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.

$$x^{2}+54x-5=500$$

Add 5 to both sides of the equation.

$$x^{2}+54x-5-\left(-5\right)=500-\left(-5\right)$$

Subtracting -5 from itself leaves 0.

$$x^{2}+54x=500-\left(-5\right)$$

Subtract -5 from 500.

$$x^{2}+54x=505$$

Divide 54, the coefficient of the x term, by 2 to get 27. Then add the square of 27 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+54x+27^{2}=505+27^{2}$$

Square 27.

$$x^{2}+54x+729=505+729$$

Add 505 to 729.

$$x^{2}+54x+729=1234$$

Factor x^{2}+54x+729. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+27\right)^{2}=1234$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+27\right)^{2}}=\sqrt{1234}$$

Simplify.

$$x+27=\sqrt{1234} x+27=-\sqrt{1234}$$

Subtract 27 from both sides of the equation.

$$x=\sqrt{1234}-27 x=-\sqrt{1234}-27$$

Solve for x

$$x=\sqrt{1234}-27\approx 8.128336141$$

$$x=-\sqrt{1234}-27\approx -62.128336141$$

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.

$$x^{2}+54x-5=500$$

Subtract 500 from both sides of the equation.

$$x^{2}+54x-5-500=500-500$$

Subtracting 500 from itself leaves 0.

$$x^{2}+54x-5-500=0$$

Subtract 500 from -5.

$$x^{2}+54x-505=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 54 for b, and -505 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-54±\sqrt{54^{2}-4\left(-505\right)}}{2}$$

Square 54.

$$x=\frac{-54±\sqrt{2916-4\left(-505\right)}}{2}$$

Multiply -4 times -505.

$$x=\frac{-54±\sqrt{2916+2020}}{2}$$

Add 2916 to 2020.

$$x=\frac{-54±\sqrt{4936}}{2}$$

Take the square root of 4936.

$$x=\frac{-54±2\sqrt{1234}}{2}$$

Now solve the equation x=\frac{-54±2\sqrt{1234}}{2} when ± is plus. Add -54 to 2\sqrt{1234}.

$$x=\frac{2\sqrt{1234}-54}{2}$$

Divide -54+2\sqrt{1234} by 2.

$$x=\sqrt{1234}-27$$

Now solve the equation x=\frac{-54±2\sqrt{1234}}{2} when ± is minus. Subtract 2\sqrt{1234} from -54.

$$x=\frac{-2\sqrt{1234}-54}{2}$$

Divide -54-2\sqrt{1234} by 2.

$$x=-\sqrt{1234}-27$$

The equation is now solved.

$$x=\sqrt{1234}-27 x=-\sqrt{1234}-27$$

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.

$$x^{2}+54x-5=500$$

Add 5 to both sides of the equation.

$$x^{2}+54x-5-\left(-5\right)=500-\left(-5\right)$$

Subtracting -5 from itself leaves 0.

$$x^{2}+54x=500-\left(-5\right)$$

Subtract -5 from 500.

$$x^{2}+54x=505$$

Divide 54, the coefficient of the x term, by 2 to get 27. Then add the square of 27 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+54x+27^{2}=505+27^{2}$$

Square 27.

$$x^{2}+54x+729=505+729$$

Add 505 to 729.

$$x^{2}+54x+729=1234$$

Factor x^{2}+54x+729. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+27\right)^{2}=1234$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+27\right)^{2}}=\sqrt{1234}$$

Simplify.

$$x+27=\sqrt{1234} x+27=-\sqrt{1234}$$

Subtract 27 from both sides of the equation.

$$x=\sqrt{1234}-27 x=-\sqrt{1234}-27$$