(3x-x^2+8)+(x^2+4x-5)=0

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Solution for (3x-x^2+8)+(x^2+4x-5)=0 equation: 



(3x-x^2+8)+(x^2+4x-5)=0

We get rid of parentheses

-x^2+x^2+3x+4x+8-5=0

We add all the numbers together, and all the variables

7x+3=0

We move all terms containing x to the left, all other terms to the right

7x=-3

x=-3/7

x=-3/7

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