How do you solve #-3x^2=2(3x-5)# using the quadratic formula?

Question

Nathan Axel · Accepted Answer

Rearrange to form: #3x^2+6x-10=0#
Substitute in quadratic formula: #x=(-(6)+-sqrt((6)^2-4(3)(-10)))/(2(3))#
Simplify to get: #x=(-3+sqrt(39))/(3) ~~1.08167# or #x=(-3-sqrt(39))/(3)~~-3.08167# First of all you need to rearrange the equation into the form #ax^2+bx+c=0# So first expand the brackets on the right hand side: #-3x^2=6x-10# Then you can either take the #-3x^2# over or the #6x-10# over. Let's just take the #-3x^2# over to give: #3x^2+6x-10=0# Now we can see what #a# #b# and #c# are: #a=3# #b=6# #c=-10# Then we substitute these into the quadratic formula: #x=(-b+-sqrt(b^2-4ac))/(2a)# Quadratics usually have 2 answers which is why there is a #+-# as for one answer you use #+# and the other #-#. So now substituting in: #x=(-(6)+-sqrt((6)^2-4(3)(-10)))/(2(3))# This simplifies to: #x=(-6+-2sqrt(39))/(6)# And dividing all terms by 2 to: #x=(-3+-sqrt(39))/(3)# which is the most simple form. So #x=(-3+sqrt(39))/(3) ~~1.08167# or #x=(-3-sqrt(39))/(3)~~-3.08167#

Eva Abramson · Answer

undefined To solve the equation -3x^2 = 2(3x - 5) using the quadratic formula, follow these steps:

1. First, rewrite the equation in standard quadratic form: -3x^2 - 6x + 10 = 0.

2. Identify the coefficients: a = -3, b = -6, and c = 10.

3. Substitute these values into the quadratic formula: x = [-b ± √(b^2 - 4ac)] / (2a).

4. Plug in the values: x = [6 ± √((-6)^2 - 4(-3)(10))] / (2 * -3).

5. Simplify inside the square root: x = [6 ± √(36 + 120)] / (-6).

6. Further simplify inside the square root: x = [6 ± √156] / (-6).

7. Now, find the two possible solutions by evaluating the square root separately for each: x = (6 + √156) / (-6) and x = (6 - √156) / (-6).

8. Lastly, simplify each expression: x = (6 ± √156) / (-6) can be simplified to x = -1 ± √(39) / 3.

So, the solutions to the equation -3x^2 = 2(3x - 5) using the quadratic formula are x = -1 + √(39) / 3 and x = -1 - √(39) / 3.