Y=0.5x^2+3x+4

Solution for Y=0.5x^2+3x+4 equation:

=0.5Y^2+3Y+4

We move all terms to the left:

-(0.5Y^2+3Y+4)=0

We get rid of parentheses

-0.5Y^2-3Y-4=0

a = -0.5; b = -3; c = -4;
Δ = b2-4ac
Δ = -32-4·(-0.5)·(-4)
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{-(-3)-1}{2*-0.5}=\frac{2}{-1}
=-2
$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+1}{2*-0.5}=\frac{4}{-1}
=-4
$