3y^2=10y+25

Solution for 3y^2=10y+25 equation:

3y^2=10y+25

We move all terms to the left:

3y^2-(10y+25)=0

We get rid of parentheses

3y^2-10y-25=0

a = 3; b = -10; c = -25;
Δ = b2-4ac
Δ = -102-4·3·(-25)
Δ = 400
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{400}=20$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-20}{2*3}=\frac{-10}{6}
=-1+2/3
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+20}{2*3}=\frac{30}{6}
=5
$