10y^2+28y+4=9y^2+39y-26

Solution for 10y^2+28y+4=9y^2+39y-26 equation:

10y^2+28y+4=9y^2+39y-26

We move all terms to the left:

10y^2+28y+4-(9y^2+39y-26)=0

We get rid of parentheses

10y^2-9y^2+28y-39y+26+4=0

We add all the numbers together, and all the variables

y^2-11y+30=0

a = 1; b = -11; c = +30;
Δ = b2-4ac
Δ = -112-4·1·30
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1}=1$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-11)-1}{2*1}=\frac{10}{2}
=5
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-11)+1}{2*1}=\frac{12}{2}
=6
$