2x^2=-8x+10

I got
x=-2-1i
x=-2+1i
But every time I plug the numbers back into the equation I am not getting the sides to equal. Not sure how to use the imaginary i.

you don't have to use the Quadratic formula for this, it is factorable upon rewriting, 2x^2+8x-10=0 and factoring out a 2, 2(x^2+4x-5)=0


You should get, 2(x+5)(x-1) making x = -5 and 1.

My chemistry professor wants us to use the quadratic formula

First you should factor out the 2 to leave X^2 + 4x -5 = 0. You should not be getting an i. Getting an i means you are a negative under the square root.

Your problem is that you're not doing the part under the square root sign correctly.
sqrt(b^2-4*a*c)
sqrt(16-4*1*-5)
sqrt(16+20)
sqrt(36)
+/- 6
so you have
(-4 +-6)/2 = -5,1
and 5,1 will satisfy the equation.

okay thank you

To solve the equation 2x^2 = -8x + 10, you can use the quadratic formula. This formula allows you to find the values of x by substituting the coefficients of the quadratic equation into the formula and simplifying the equation.

First, rearrange the equation to bring all terms to one side:

2x^2 + 8x - 10 = 0

Next, identify the coefficients a, b, and c:

a = 2
b = 8
c = -10

Now, substitute these values into the quadratic formula:

x = (-b ± √(b^2 - 4ac)) / 2a

Plugging in the values, we get:

x = (-8 ± √(8^2 - 4*2*(-10))) / (2*2)

Simplifying the expression under the square root:

x = (-8 ± √(64 + 80)) / 4

x = (-8 ± √(144)) / 4

x = (-8 ± 12) / 4

Now, consider both the positive and negative square root solutions:

For the positive square root:

x = (-8 + 12) / 4
x = 4 / 4
x = 1

For the negative square root:

x = (-8 - 12) / 4
x = -20 / 4
x = -5

Thus, the two solutions to the equation 2x^2 = -8x + 10 are x = 1 and x = -5.

Now, let's substitute these solutions back into the original equation to verify if they satisfy the equation:

For x = 1:
2(1)^2 = -8(1) + 10
2 = -8 + 10
2 = 2 (LHS = RHS)

For x = -5:
2(-5)^2 = -8(-5) + 10
2(25) = 40 + 10
50 = 50 (LHS = RHS)

As you can see, both solutions satisfy the original equation, so x = 1 and x = -5 are indeed the correct solutions.

Now, regarding the use of the imaginary unit i, in this specific equation, there are no complex solutions. The solutions obtained are real numbers. However, if you encounter an equation where the discriminant (b^2 - 4ac) is negative, it means the quadratic equation has complex solutions involving the imaginary unit i.