Solve the quadratic by factoring. x^(2)-3x-77=-5x+3 Answer: x=

Move Terms, Set Equal: First, we need to move all terms to one side of the equation to set it equal to zero. x^2 - 3x - 77 = -5x + 3 Add 5x to both sides and subtract 3 from both sides to get: x^2 + 2x - 80 = 0Factor Quadratic Equation: Now, we need to factor the quadratic equation x^2 + 2x - 80. We are looking for two numbers that multiply to -80 and add up to 2. The numbers 10 and -8 satisfy these conditions because 10 * -8 = -80 and 10 + (-8) = 2. So we can write the factored form as: (x + 10)(x - 8) = 0Solve for First Factor: Next, we set each factor equal to zero and solve for x. First factor: x + 10 = 0 Subtract 10 from both sides to solve for x: x = -10Solve for Second Factor: Now, solve for x using the second factor: x - 8 = 0 Add 8 to both sides to solve for x: x = 8

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Algebra 1

Factor quadratics with leading coefficient 1

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Q. Solve the quadratic by factoring.

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x^{2}-3 x-77=-5 x+3

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Answer:

x=

Move Terms, Set Equal: First, we need to move all terms to one side of the equation to set it equal to zero. $\newline$ $x^2 - 3x - 77 = -5x + 3$ $\newline$ Add $5x$ to both sides and subtract $3$ from both sides to get: $\newline$ $x^2 + 2x - 80 = 0$
Factor Quadratic Equation: Now, we need to factor the quadratic equation $x^2 + 2x - 80$ . We are looking for two numbers that multiply to $-80$ and add up to $2$ . The numbers $10$ and $-8$ satisfy these conditions because $10 \times -8 = -80$ and $10 + (-8) = 2$ . So we can write the factored form as: $(x + 10)(x - 8) = 0$
Solve for First Factor: Next, we set each factor equal to zero and solve for $x$ . $\newline$ First factor: $x + 10 = 0$ $\newline$ Subtract $10$ from both sides to solve for $x$ : $\newline$ $x = -10$
Solve for Second Factor: Now, solve for $x$ using the second factor: $x - 8 = 0$ Add $8$ to both sides to solve for $x$ : $x = 8$