Find one value of x that is a solution to the equation: (2x+3)^(2)-4(2x+3)-12=0 x=

Identifying the quadratic equation: Let's first identify the quadratic equation in the form of a single variable expression by setting u = 2x + 3. The equation becomes: u² - 4u - 12 = 0Factoring the quadratic equation: Now, we need to factor the quadratic equation u² - 4u - 12 = 0. We look for two numbers that multiply to -12 and add up to -4. The numbers -6 and +2 fit these requirements. So we can write the equation as: (u - 6)(u + 2) = 0Solving for u: Next, we solve for u by setting each factor equal to zero: u - 6 = 0 or u + 2 = 0 This gives us two possible solutions for u: u = 6 or u = -2Substituting back for u: We now substitute back for u with 2x + 3 to find the values of x: 2x + 3 = 6 or 2x + 3 = -2Solving the first equation for x: Solving the first equation 2x + 3 = 6 for x: 2x = 6 - 3 2x = 3 x = 3/2Solving the second equation for x: Solving the second equation 2x + 3 = -2 for x: 2x = -2 - 3 2x = -5 x = -5/2Final solutions for x: We have found two values for x that satisfy the original equation: x = 3/2 or x = -5/2 We can choose either one as a correct solution to the original problem.

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Find one value of
x that is a solution to the equation:

(2x+3)^(2)-4(2x+3)-12=0

x=

Find one value of $x$ that is a solution to the equation: $\newline$ $(2x+3)^{2}-4(2x+3)-12=0$ $\newline$ $x=$

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Math Problems

Algebra 1

Solve a quadratic equation by factoring

Full solution

Q. Find one value of

x

that is a solution to the equation:

\newline

(2x+3)^{2}-4(2x+3)-12=0

\newline

x=

Identifying the quadratic equation: Let's first identify the quadratic equation in the form of a single variable expression by setting $u = 2x + 3$ . $\newline$ The equation becomes: $\newline$ $u^2 - 4u - 12 = 0$
Factoring the quadratic equation: Now, we need to factor the quadratic equation $u^2 - 4u - 12 = 0$ . $\newline$ We look for two numbers that multiply to $-12$ and add up to $-4$ . The numbers $-6$ and $+2$ fit these requirements. $\newline$ So we can write the equation as: $\newline$ $(u - 6)(u + 2) = 0$
Solving for u: Next, we solve for u by setting each factor equal to zero: $\newline$ $u - 6 = 0$ or $u + 2 = 0$ $\newline$ This gives us two possible solutions for u: $\newline$ $u = 6$ or $u = -2$
Substituting back for u: We now substitute back for u with $2x + 3$ to find the values of x: $\newline$ $2x + 3 = 6$ or $2x + 3 = -2$
Solving the first equation for x: Solving the first equation $2x + 3 = 6$ for $x$ : $\newline$ $2x = 6 - 3$ $\newline$ $2x = 3$ $\newline$ $x = \frac{3}{2}$
Solving the second equation for x: Solving the second equation $2x + 3 = -2$ for x: $\newline$ $2x = -2 - 3$ $\newline$ $2x = -5$ $\newline$ $x = -\frac{5}{2}$
Final solutions for x: We have found two values for x that satisfy the original equation: $\newline$ x = $\frac{3}{2}$ or x = $-\frac{5}{2}$ $\newline$ We can choose either one as a correct solution to the original problem.