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

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Answer 1 · 2015-10-02T13:50:34

$x=6$

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

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