Question

select all the correct answers.
what are the solutions to this equation?
$2x^2 = -10x + 12$
$\square\\ x = 3$
$\square\\ x = -3$
$\square\\ x = 1$
$\square\\ x = 6$
$\square\\ x = -6$
$\square\\ x = -2$

Explanation:

Step1: Rewrite the equation

First, we rewrite the given equation \(2x^{2}=-10x + 12\) in standard quadratic form \(ax^{2}+bx + c = 0\). We add \(10x\) and subtract \(12\) from both sides:
\(2x^{2}+10x - 12=0\)
We can simplify this equation by dividing all terms by \(2\):
\(x^{2}+5x - 6 = 0\)

Step2: Factor the quadratic equation

We factor the quadratic equation \(x^{2}+5x - 6=0\). We need two numbers that multiply to \(- 6\) and add up to \(5\). The numbers are \(6\) and \(-1\) since \(6\times(-1)=-6\) and \(6+( - 1)=5\). So we can factor the equation as:
\((x + 6)(x - 1)=0\)

Step3: Solve for \(x\)

Using the zero - product property, if \(ab = 0\), then either \(a = 0\) or \(b = 0\).
For \((x + 6)(x - 1)=0\), we set each factor equal to zero:

• If \(x+6 = 0\), then \(x=-6\).
• If \(x - 1=0\), then \(x = 1\).

Wait, we made a mistake in factoring. Let's go back to the equation \(2x^{2}+10x - 12 = 0\) or the original equation \(2x^{2}=-10x + 12\). Let's use the quadratic formula or re - factor correctly.

Starting over from the original equation \(2x^{2}+10x-12 = 0\), divide by \(2\) to get \(x^{2}+5x - 6=0\). Wait, actually, let's use the quadratic formula for \(ax^{2}+bx + c = 0\), \(x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\). For \(2x^{2}+10x - 12=0\), \(a = 2\), \(b = 10\), \(c=-12\).

First, calculate the discriminant \(\Delta=b^{2}-4ac=(10)^{2}-4\times2\times(-12)=100 + 96=196\)

Then \(x=\frac{-10\pm\sqrt{196}}{2\times2}=\frac{-10\pm14}{4}\)

Case 1: \(x=\frac{-10 + 14}{4}=\frac{4}{4}=1\)

Case 2: \(x=\frac{-10-14}{4}=\frac{-24}{4}=-6\)

Wait, but let's check the answer by plugging back into the original equation.

For \(x = 1\): Left - hand side (LHS) \(=2\times(1)^{2}=2\), Right - hand side (RHS) \(=-10\times1 + 12=-10 + 12 = 2\). So \(x = 1\) is a solution.

For \(x=-6\): LHS \(=2\times(-6)^{2}=2\times36 = 72\), RHS \(=-10\times(-6)+12=60 + 12 = 72\). So \(x=-6\) is a solution.

Wait, but let's try another approach. Let's go back to the original equation \(2x^{2}+10x - 12=0\), divide by \(2\) to get \(x^{2}+5x - 6 = 0\). Wait, maybe I made a mistake in the factoring step earlier. Let's try to factor \(x^{2}+5x - 6\) again. We need two numbers that multiply to \(-6\) and add to \(5\). The numbers are \(6\) and \(-1\) because \(6\times(-1)=-6\) and \(6+( - 1)=5\). So \((x + 6)(x - 1)=0\), which gives \(x=-6\) or \(x = 1\).

But let's check the options given. The options are \(x = 3\), \(x=-3\), \(x = 1\), \(x = 6\), \(x=-6\), \(x=-2\).

Wait, maybe I made a mistake in the initial rearrangement. Let's start from the original equation \(2x^{2}=-10x + 12\). Let's move all terms to the left: \(2x^{2}+10x-12 = 0\). Let's divide by \(2\): \(x^{2}+5x - 6=0\). Wait, let's use the quadratic formula correctly. \(a = 1\), \(b = 5\), \(c=-6\). Then \(x=\frac{-5\pm\sqrt{25+24}}{2}=\frac{-5\pm\sqrt{49}}{2}=\frac{-5\pm7}{2}\)

When we take the plus sign: \(\frac{-5 + 7}{2}=\frac{2}{2}=1\)

When we take the minus sign: \(\frac{-5-7}{2}=\frac{-12}{2}=-6\)

So the solutions are \(x = 1\) and \(x=-6\)

Answer:

C. \(x = 1\), E. \(x=-6\) (where we assume the options are labeled as follows: A. \(x = 3\), B. \(x=-3\), C. \(x = 1\), D. \(x = 6\), E. \(x=-6\), F. \(x=-2\))