-3=12y(-10y+35)

Solution for -3=12y(-10y+35) equation:

-3=12y(-10y+35)

We move all terms to the left:

-3-(12y(-10y+35))=0


We calculate terms in parentheses: -(12y(-10y+35)), so:

12y(-10y+35)

We multiply parentheses

-120y^2+420y

Back to the equation:

-(-120y^2+420y)


We get rid of parentheses

120y^2-420y-3=0

a = 120; b = -420; c = -3;
Δ = b2-4ac
Δ = -4202-4·120·(-3)
Δ = 177840
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{177840}=\sqrt{144*1235}=\sqrt{144}*\sqrt{1235}=12\sqrt{1235}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-420)-12\sqrt{1235}}{2*120}=\frac{420-12\sqrt{1235}}{240}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-420)+12\sqrt{1235}}{2*120}=\frac{420+12\sqrt{1235}}{240}
$