2x^2+3x-4.5=0

Solution for 2x^2+3x-4.5=0 equation:

2x^2+3x-4.5=0

a = 2; b = 3; c = -4.5;
Δ = b2-4ac
Δ = 32-4·2·(-4.5)
Δ = 45
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:

$\sqrt{\Delta}=\sqrt{45}=\sqrt{9*5}=\sqrt{9}*\sqrt{5}=3\sqrt{5}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-3\sqrt{5}}{2*2}=\frac{-3-3\sqrt{5}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+3\sqrt{5}}{2*2}=\frac{-3+3\sqrt{5}}{4}
$