0.5x^2+10x=16

Solution for 0.5x^2+10x=16 equation:

0.5x^2+10x=16

We move all terms to the left:

0.5x^2+10x-(16)=0

a = 0.5; b = 10; c = -16;
Δ = b2-4ac
Δ = 102-4·0.5·(-16)
Δ = 132
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{132}=\sqrt{4*33}=\sqrt{4}*\sqrt{33}=2\sqrt{33}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-2\sqrt{33}}{2*0.5}=\frac{-10-2\sqrt{33}}{1}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+2\sqrt{33}}{2*0.5}=\frac{-10+2\sqrt{33}}{1}
$