Algebraic Equations and Expressions

We can solve this equation using the quadratic formula: $x = (-b ± √(b^2 - 4ac))/(2a)$. Substituting the given values, we get $x = (-(-5) ± √((-5)^2 - 4(2)(3)))/(2(2))$. Simplifying, we get $x = (5 ± √(25 - 24))/(4) = (5 ± 1)/(4) = 1/2$ or $x = 2$.

To solve this equation, we can subtract 5 from both sides: $3x + 5 - 5 = 17 - 5$. Simplifying, we get $3x = 12$. Dividing both sides by 3, we get $x = 4$.

To solve the equation 2x^2 + 3x - 5 = 0, we can use the quadratic formula, which is given by the formula x = (-b ± √(b^2 - 4ac)) / 2a. In this case, a = 2, b = 3, and c = -5, so x = (-3 ± √(3^2 - 4(2)(-5))) / 2(2) = (-3 ± √(9 + 40)) / 4 = (-3 ± √49) / 4 = (-3 ± 7) / 4. Therefore, x = 1 or x = -2.

Using the quadratic formula, x = (-b ± √(b² - 4ac)) / 2a, where a = 1, b = -5, and c = 6, we get x = (5 ± √(25 - 24)) / 2 = (5 ± 1) / 2. Therefore, x = 2 or x = 3.

To solve for x, we need to isolate x on one side of the equation. We can do this by subtracting 5 from both sides of the equation: 3x + 5 - 5 = 17 - 5. This simplifies to 3x = 12. Then, we divide both sides of the equation by 3: 3x / 3 = 12 / 3. This simplifies to x = 4.

To solve for x, we can subtract 5 from both sides of the equation: 3x + 5 - 5 = 17 - 5, which simplifies to 3x = 12. Dividing both sides by 3, we get x = 4.

First, evaluate the parentheses: (3 + 4) = 7 and (5 - 2) = 3. Then, perform the multiplication: 7 x 3 = 21. Therefore, the value of the expression is 21.

Using the quadratic formula, (x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}), we can solve for (x) in the equation (x^2 - 2x + 1 = 0). Substituting the values of (a), (b), and (c), we get (x = 1).

Using the rational root theorem, we can find that (x = 1) is a root of the equation (x^3 - 3x^2 + 3x - 1 = 0). Therefore, (x - 1) is a factor of the polynomial. Dividing the polynomial by (x - 1), we get (x^2 - 2x + 1). Solving for (x) in this quadratic equation, we get (x = 1).

Using the rational root theorem, we can find that (x = 1) is a root of the equation (x^4 - 2x^3 + x^2 - 2x + 1 = 0). Therefore, (x - 1) is a factor of the polynomial. Dividing the polynomial by (x - 1), we get (x^3 - x^2 + 1). Solving for (x) in this cubic equation, we get (x = 1).

Using the rational root theorem, we can find that (x = 1) is a root of the equation (x^5 - 3x^4 + 3x^3 - x^2 + x - 1 = 0). Therefore, (x - 1) is a factor of the polynomial. Dividing the polynomial by (x - 1), we get (x^4 - 2x^3 + 2x^2 - x + 1). Solving for (x) in this quartic equation, we get (x = 1).

Using the rational root theorem, we can find that (x = 1) is a root of the equation (x^6 - 4x^5 + 6x^4 - 4x^3 + x^2 - 4x + 1 = 0). Therefore, (x - 1) is a factor of the polynomial. Dividing the polynomial by (x - 1), we get (x^5 - 3x^4 + 3x^3 - x^2 + x - 1). Solving for (x) in this quintic equation, we get (x = 1).

Using the rational root theorem, we can find that (x = 1) is a root of the equation (x^7 - 5x^6 + 10x^5 - 10x^4 + 5x^3 - x^2 + x - 1 = 0). Therefore, (x - 1) is a factor of the polynomial. Dividing the polynomial by (x - 1), we get (x^6 - 4x^5 + 6x^4 - 4x^3 + x^2 - 4x + 1). Solving for (x) in this sextic equation, we get (x = 1).

The Hypergeometric function (_2F_1(1, 2; 3; x)) at x = 0 is equal to 1.

The Hypergeometric function (_2F_1(2, 3; 4; x)) at x = 1 is equal to 1.