0.5x^2+10x+8=0

Solution for 0.5x^2+10x+8=0 equation:

0.5x^2+10x+8=0

a = 0.5; b = 10; c = +8;
Δ = b2-4ac
Δ = 102-4·0.5·8
Δ = 84
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:

$\sqrt{\Delta}=\sqrt{84}=\sqrt{4*21}=\sqrt{4}*\sqrt{21}=2\sqrt{21}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-2\sqrt{21}}{2*0.5}=\frac{-10-2\sqrt{21}}{1}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+2\sqrt{21}}{2*0.5}=\frac{-10+2\sqrt{21}}{1}
$