How do you solve $4+7(3x-5)^2 =1$ ?

Question

1372 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-06-08T15:03:53

$x_1=1/21(35-isqrt(21))$
$x_2=1/21(35+isqrt(21))$

Expanding after the frormula

$(a-b)^2=a^2-2ab+b^2$
we get
$63x^2-210x+148=0$
dividing all by $63$

$x^2-210/63x+148/63=0$
solving this equation

$x_1=1/21(35-isqrt(21))$

$x_2=1/21(35+isqrt(21))$

where $i^2=-1$

How do you solve 4+7(3x-5)^2 =14+7(3x-5)^2 =1?