4x^2-18x+2.5=0

Solution for 4x^2-18x+2.5=0 equation:

4x^2-18x+2.5=0

a = 4; b = -18; c = +2.5;
Δ = b2-4ac
Δ = -182-4·4·2.5
Δ = 284
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:

$\sqrt{\Delta}=\sqrt{284}=\sqrt{4*71}=\sqrt{4}*\sqrt{71}=2\sqrt{71}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-18)-2\sqrt{71}}{2*4}=\frac{18-2\sqrt{71}}{8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-18)+2\sqrt{71}}{2*4}=\frac{18+2\sqrt{71}}{8}
$