0.5x^2+18x-64=0

Solution for 0.5x^2+18x-64=0 equation:

0.5x^2+18x-64=0

a = 0.5; b = 18; c = -64;
Δ = b2-4ac
Δ = 182-4·0.5·(-64)
Δ = 452
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:

$\sqrt{\Delta}=\sqrt{452}=\sqrt{4*113}=\sqrt{4}*\sqrt{113}=2\sqrt{113}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(18)-2\sqrt{113}}{2*0.5}=\frac{-18-2\sqrt{113}}{1}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(18)+2\sqrt{113}}{2*0.5}=\frac{-18+2\sqrt{113}}{1}
$