0,5x2-9x+16=0

Solution for 0,5x2-9x+16=0 equation:

0.5x^2-9x+16=0

a = 0.5; b = -9; c = +16;
Δ = b2-4ac
Δ = -92-4·0.5·16
Δ = 49
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{49}=7$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-9)-7}{2*0.5}=\frac{2}{1}
=2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-9)+7}{2*0.5}=\frac{16}{1}
=16
$