10x-0.5x^2-40=0

Solution for 10x-0.5x^2-40=0 equation:

10x-0.5x^2-40=0

a = -0.5; b = 10; c = -40;
Δ = b2-4ac
Δ = 102-4·(-0.5)·(-40)
Δ = 20
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:

$\sqrt{\Delta}=\sqrt{20}=\sqrt{4*5}=\sqrt{4}*\sqrt{5}=2\sqrt{5}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-2\sqrt{5}}{2*-0.5}=\frac{-10-2\sqrt{5}}{-1}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+2\sqrt{5}}{2*-0.5}=\frac{-10+2\sqrt{5}}{-1}
$