0.5x2+x=40

Solution for 0.5x2+x=40 equation:

0.5x^2+x=40

We move all terms to the left:

0.5x^2+x-(40)=0

a = 0.5; b = 1; c = -40;
Δ = b2-4ac
Δ = 12-4·0.5·(-40)
Δ = 81
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{81}=9$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-9}{2*0.5}=\frac{-10}{1}
=-10
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+9}{2*0.5}=\frac{8}{1}
=8
$