21y^2=62y^+3

Solution for 21y^2=62y^+3 equation:

21y^2=62y^+3

We move all terms to the left:

21y^2-(62y^+3)=0

We get rid of parentheses

21y^2-62y^-3=0

We add all the numbers together, and all the variables

21y^2-62y-3=0

a = 21; b = -62; c = -3;
Δ = b2-4ac
Δ = -622-4·21·(-3)
Δ = 4096
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4096}=64$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-62)-64}{2*21}=\frac{-2}{42}
=-1/21
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-62)+64}{2*21}=\frac{126}{42}
=3
$