Solve for x. x^2 - 6x = 5

Rephrase the Problem: First, let's rephrase the problem into a ＂What are the values of x that satisfy the equation x squared minus 6x equals 5?＂Move Terms: To solve the equation x^2 - 6x = 5, we need to move all terms to one side to set the equation equal to zero. We do this by subtracting 5 from both sides of the equation. x^2 - 6x - 5 = 0Quadratic Formula: Now we have a quadratic equation in the standard form. We can attempt to factor it, or use the quadratic formula to find the values of x. The quadratic formula is x = (-b ± √(b^2 - 4ac)) / (2a), where a, b, and c are the coefficients from the quadratic equation ax^2 + bx + c = 0. In our case, a = 1, b = -6, and c = -5.Calculate Discriminant: Let's calculate the discriminant (b^2 - 4ac) to see if factoring is possible or if we will have real solutions. Discriminant = (-6)^2 - 4(1)(-5) = 36 + 20 = 56 Since the discriminant is positive, we have two real solutions, and it suggests that the equation might be factorable. However, since 56 is not a perfect square, factoring might be difficult or impossible. We will use the quadratic formula.Apply Quadratic Formula: Applying the quadratic formula, we get: x = (-(-6) ± √(56)) / (2*1) x = (6 ± √(56)) / 2 x = (6 ± √(4*14)) / 2 x = (6 ± 2√(14)) / 2Simplify the Expression: We can simplify the expression by dividing both terms in the numerator by 2: x = 3 ± √(14) So we have two solutions for x.

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Solve for $x$ . $\newline$ $x^2 - 6x = 5$

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\newline

x^2 - 6x = 5

Rephrase the Problem: First, let's rephrase the problem into a ＂What are the values of $x$ that satisfy the equation $x^2 - 6x = 5$ ?＂
Move Terms: To solve the equation $x^2 - 6x = 5$ , we need to move all terms to one side to set the equation equal to zero. We do this by subtracting $5$ from both sides of the equation. $\newline$ $x^2 - 6x - 5 = 0$
Quadratic Formula: Now we have a quadratic equation in the standard form. We can attempt to factor it, or use the quadratic formula to find the values of $x$ . The quadratic formula is $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ , where $a$ , $b$ , and $c$ are the coefficients from the quadratic equation $ax^2 + bx + c = 0$ . In our case, $a = 1$ , $b = -6$ , and $c = -5$ .
Calculate Discriminant: Let's calculate the discriminant $(b^2 - 4ac)$ to see if factoring is possible or if we will have real solutions. $\newline$ Discriminant = $(-6)^2 - 4(1)(-5) = 36 + 20 = 56$ $\newline$ Since the discriminant is positive, we have two real solutions, and it suggests that the equation might be factorable. However, since $56$ is not a perfect square, factoring might be difficult or impossible. We will use the quadratic formula.
Apply Quadratic Formula: Applying the quadratic formula, we get: $\newline$ $x = \frac{-(-6) \pm \sqrt{56}}{2 \cdot 1}$ $\newline$ $x = \frac{6 \pm \sqrt{56}}{2}$ $\newline$ $x = \frac{6 \pm \sqrt{4 \cdot 14}}{2}$ $\newline$ $x = \frac{6 \pm 2\sqrt{14}}{2}$
Simplify the Expression: We can simplify the expression by dividing both terms in the numerator by $2$ : $x = 3 \pm \sqrt{(14)}$ So we have two solutions for $x$ .