x^2-5x+3y=20-এর সমাধান করুন

$$x ^ { 2 } - 5 x + 3 y = 20$$

Solve for x

$$x=\frac{\sqrt{105-12y}+5}{2}$$

$$x=\frac{-\sqrt{105-12y}+5}{2}\text{, }y\leq \frac{35}{4}$$

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.

$$x^{2}-5x+3y=20$$

Subtract 20 from both sides of the equation.

$$x^{2}-5x+3y-20=20-20$$

Subtracting 20 from itself leaves 0.

$$x^{2}-5x+3y-20=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -5 for b, and 3y-20 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-5\right)±\sqrt{\left(-5\right)^{2}-4\left(3y-20\right)}}{2}$$

Square -5.

$$x=\frac{-\left(-5\right)±\sqrt{25-4\left(3y-20\right)}}{2}$$

Multiply -4 times 3y-20.

$$x=\frac{-\left(-5\right)±\sqrt{25+80-12y}}{2}$$

Add 25 to -12y+80.

$$x=\frac{-\left(-5\right)±\sqrt{105-12y}}{2}$$

The opposite of -5 is 5.

$$x=\frac{5±\sqrt{105-12y}}{2}$$

Now solve the equation x=\frac{5±\sqrt{105-12y}}{2} when ± is plus. Add 5 to \sqrt{105-12y}.

$$x=\frac{\sqrt{105-12y}+5}{2}$$

Now solve the equation x=\frac{5±\sqrt{105-12y}}{2} when ± is minus. Subtract \sqrt{105-12y} from 5.

$$x=\frac{-\sqrt{105-12y}+5}{2}$$

The equation is now solved.

$$x=\frac{\sqrt{105-12y}+5}{2} x=\frac{-\sqrt{105-12y}+5}{2}$$

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.

$$x^{2}-5x+3y=20$$

Subtract 3y from both sides of the equation.

$$x^{2}-5x+3y-3y=20-3y$$

Subtracting 3y from itself leaves 0.

$$x^{2}-5x=20-3y$$

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

$$x^{2}-5x+\left(-\frac{5}{2}\right)^{2}=20-3y+\left(-\frac{5}{2}\right)^{2}$$

Square -\frac{5}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-5x+\frac{25}{4}=20-3y+\frac{25}{4}$$

Add 20-3y to \frac{25}{4}.

$$x^{2}-5x+\frac{25}{4}=\frac{105}{4}-3y$$

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

$$\left(x-\frac{5}{2}\right)^{2}=\frac{105}{4}-3y$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{5}{2}\right)^{2}}=\sqrt{\frac{105}{4}-3y}$$

Simplify.

$$x-\frac{5}{2}=\frac{\sqrt{105-12y}}{2} x-\frac{5}{2}=-\frac{\sqrt{105-12y}}{2}$$

Add \frac{5}{2} to both sides of the equation.

$$x=\frac{\sqrt{105-12y}+5}{2} x=\frac{-\sqrt{105-12y}+5}{2}$$

Solve for y

$$y=\frac{20+5x-x^{2}}{3}$$

Steps for Solving Linear Equation

Subtract x^{2} from both sides.

$$-5x+3y=20-x^{2}$$

Add 5x to both sides.

$$3y=20-x^{2}+5x$$

The equation is in standard form.

$$3y=20+5x-x^{2}$$

Divide both sides by 3.

$$\frac{3y}{3}=\frac{20+5x-x^{2}}{3}$$

Dividing by 3 undoes the multiplication by 3.

$$y=\frac{20+5x-x^{2}}{3}$$