3y^2=15y+42

Solution for 3y^2=15y+42 equation:

3y^2=15y+42

We move all terms to the left:

3y^2-(15y+42)=0

We get rid of parentheses

3y^2-15y-42=0

a = 3; b = -15; c = -42;
Δ = b2-4ac
Δ = -152-4·3·(-42)
Δ = 729
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{729}=27$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-15)-27}{2*3}=\frac{-12}{6}
=-2
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-15)+27}{2*3}=\frac{42}{6}
=7
$