(50y^2)-(350y)+(600)=0

Solution for (50y^2)-(350y)+(600)=0 equation:

(50y^2)-(350y)+(600)=0

a = 50; b = -350; c = +600;
Δ = b2-4ac
Δ = -3502-4·50·600
Δ = 2500
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{2500}=50$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-350)-50}{2*50}=\frac{300}{100}
=3
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-350)+50}{2*50}=\frac{400}{100}
=4
$