How do you solve $3-13s=3-8s+s^2$ ?

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Answer 1 · 2016-04-11T09:17:10

$s=-5$

Our aim is to isolate and find the value of $s$

$color(blue)(3-13s=3-8s+s^2$

Subtract $3$ both sides

$rarrcolor(blue)(3)-13s-color(blue)(3)=color(blue)(3)-8s+s^2-color(blue)(3)$

$rarr-13s=-8s+s^2$

Add $8s$ both sides

$rarrcolor(blue)(-13s)+color(blue)(8s)=color(blue)(-8s)+s^2+color(blue)(-8)$

$rarr-5s=s^2$

From now use logic!

We know that $color(brown)(s^2=s*s$

So,

$rarr-5s=s*s$

Divide both sides by $s$

$rarr(-5cancelcolor(blue)(s))/cancelcolor(blue)(s)=(s*cancelcolor(blue)(s))/cancelcolor(blue)(s)$

$color(green)(rArr-5=s$

How do you solve 3-13s=3-8s+s^23-13s=3-8s+s^2?