10y^2+33y-3=0

Solution for 10y^2+33y-3=0 equation:

10y^2+33y-3=0

a = 10; b = 33; c = -3;
Δ = b2-4ac
Δ = 332-4·10·(-3)
Δ = 1209
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(33)-\sqrt{1209}}{2*10}=\frac{-33-\sqrt{1209}}{20}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(33)+\sqrt{1209}}{2*10}=\frac{-33+\sqrt{1209}}{20}
$