-3=12-5y(2y-7)

Solution for -3=12-5y(2y-7) equation:

-3=12-5y(2y-7)

We move all terms to the left:

-3-(12-5y(2y-7))=0


We calculate terms in parentheses: -(12-5y(2y-7)), so:

12-5y(2y-7)

determiningTheFunctionDomain
-5y(2y-7)+12

We multiply parentheses

-10y^2+35y+12

Back to the equation:

-(-10y^2+35y+12)


We get rid of parentheses

10y^2-35y-12-3=0

We add all the numbers together, and all the variables

10y^2-35y-15=0

a = 10; b = -35; c = -15;
Δ = b2-4ac
Δ = -352-4·10·(-15)
Δ = 1825
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:

$\sqrt{\Delta}=\sqrt{1825}=\sqrt{25*73}=\sqrt{25}*\sqrt{73}=5\sqrt{73}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-35)-5\sqrt{73}}{2*10}=\frac{35-5\sqrt{73}}{20}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-35)+5\sqrt{73}}{2*10}=\frac{35+5\sqrt{73}}{20}
$