2x^2+16x-1.5=0

Solution for 2x^2+16x-1.5=0 equation:

2x^2+16x-1.5=0

a = 2; b = 16; c = -1.5;
Δ = b2-4ac
Δ = 162-4·2·(-1.5)
Δ = 268
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:

$\sqrt{\Delta}=\sqrt{268}=\sqrt{4*67}=\sqrt{4}*\sqrt{67}=2\sqrt{67}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-2\sqrt{67}}{2*2}=\frac{-16-2\sqrt{67}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+2\sqrt{67}}{2*2}=\frac{-16+2\sqrt{67}}{4}
$