How do you solve $\frac { 3} { 2} ( x - 10) ^ { 2} = \frac { 1} { 2}$ ?

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Answer 1 · 2017-07-22T14:07:17

$x={(30+sqrt3)/3 , (30-sqrt3)/3}$

$3/2(x-10)^2=1/2$
Multiply both by 2
$3(x-10)^2=1$
$3(x^2-20+100)=1$
$3x^2-60+300=1$
$3x^2-60+299=0$

We know that if $ax^2+bx+c=0$ then $x=(-b+-sqrt(b^2-4ac))/(2a)$

So $x=(-(-60)+-sqrt(60^2-4*3*299))/(2*3)$
$x=(60+-sqrt(3600-3588))/6$
$x=(60+-sqrt(12))/6$
$x=(60+-2sqrt3)/6$
$x=(30+-sqrt3)/3$

So, $x=(30+sqrt3)/3$ or $x=(30-sqrt3)/3$

How do you solve \frac { 3} { 2} ( x - 10) ^ { 2} = \frac { 1} { 2}\frac { 3} { 2} ( x - 10) ^ { 2} = \frac { 1} { 2}?