How do you solve #3(x^2-2x+5) -2(3x-5) = 4(x^2-x-5)-(x^2+3)#?

Answer 1

Isaiah Anthony

#x=6#

First, expand each expression. #3(x^2−2x+5)−2(3x−5)=4(x^2−x−5)−(x^2+3)# #3x^2−6x+15−6x+10=4x^2−4x−20−x^2-3#

Next, combine everything in one side. #3x^2−6x+15−6x+10=4x^2−4x−20−x^2-3# #3x^2−6x+15−6x+10-4x^2+4x+20+x^2+3=0#

Combine all like terms. #3x^2−6x+15−6x+10-4x^2+4x+20+x^2+3=0# #3x^2-4x^2+x^2-6x-6x+4x+15+10+20+3=0# #-8x+48=0#

Isolate #x#. #-8x+48=0# #-8x=-48# #color(red)(x=6)#

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Answer 2

Eva Abramson

To solve the equation (3(x^2-2x+5) - 2(3x-5) = 4(x^2-x-5) - (x^2+3)):

Expand and simplify each side of the equation.
Combine like terms.
Move all terms to one side to set the equation to zero.
Factor the quadratic expression.
Solve for (x) by factoring or using the quadratic formula.

After following these steps, you'll find the value(s) of (x) that satisfy the equation.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.