11.3y^2+20y-23=0

Solution for 11.3y^2+20y-23=0 equation:

11.3y^2+20y-23=0

a = 11.3; b = 20; c = -23;
Δ = b2-4ac
Δ = 202-4·11.3·(-23)
Δ = 1439.6
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{-(20)-\sqrt{1439.6}}{2*11.3}=\frac{-20-\sqrt{1439.6}}{22.6}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+\sqrt{1439.6}}{2*11.3}=\frac{-20+\sqrt{1439.6}}{22.6}
$