y^2+3y-3.75=0

Solution for y^2+3y-3.75=0 equation:

y^2+3y-3.75=0

a = 1; b = 3; c = -3.75;
Δ = b2-4ac
Δ = 32-4·1·(-3.75)
Δ = 24
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{24}=\sqrt{4*6}=\sqrt{4}*\sqrt{6}=2\sqrt{6}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-2\sqrt{6}}{2*1}=\frac{-3-2\sqrt{6}}{2}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+2\sqrt{6}}{2*1}=\frac{-3+2\sqrt{6}}{2}
$