$4x^2+72x+320=0$ from least to greatest

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Grade 8

Evaluate expressions using properties of exponents

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4x^2+72x+320=0

from least to greatest

Identify the quadratic equation: Identify the quadratic equation to be solved. $\newline$ The given quadratic equation is $4x^2 + 72x + 320 = 0$ . We need to find the values of $x$ that satisfy this equation.
Factor the quadratic equation: Factor the quadratic equation if possible. $\newline$ We look for two numbers that multiply to give the product of the coefficient of $x^2$ term ( $4$ ) and the constant term ( $320$ ), and add up to the coefficient of the $x$ term ( $72$ ). $\newline$ The two numbers that satisfy these conditions are $40$ and $8$ , since $4\times320 = 1280$ and $40 + 8 = 48$ . However, we need a sum of $72$ , not $4$ $0$ , so factoring might not be straightforward. We should try another method.
Use the quadratic formula: Use the quadratic formula to solve for $x$ . 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 $ax^2 + bx + c = 0$ . For our equation, $a = 4$ , $b = 72$ , and $c = 320$ .
Calculate the discriminant: Calculate the discriminant $b^2 - 4ac$ . The discriminant is $72^2 - 4\cdot4\cdot320$ . Calculating the discriminant: $72^2 = 5184$ and $4\cdot4\cdot320 = 5120$ . So, the discriminant is $5184 - 5120 = 64$ .
Calculate the two solutions: Calculate the two solutions using the quadratic formula. $\newline$ First solution: $x = \frac{-72 + \sqrt{64}}{2 \times 4}$ $\newline$ Second solution: $x = \frac{-72 - \sqrt{64}}{2 \times 4}$ $\newline$ $\sqrt{64} = 8$ , so we have: $\newline$ First solution: $x = \frac{-72 + 8}{8}$ $\newline$ Second solution: $x = \frac{-72 - 8}{8}$
Simplify the solutions: Simplify the solutions. $\newline$ First solution: $x = (-64) / 8 = -8$ $\newline$ Second solution: $x = (-80) / 8 = -10$
Arrange the solutions: Arrange the solutions in ascending order. $\newline$ The solutions in ascending order are $-10$ and $-8$ .