8x^2-12x+3.5=0

Solution for 8x^2-12x+3.5=0 equation:

8x^2-12x+3.5=0

a = 8; b = -12; c = +3.5;
Δ = b2-4ac
Δ = -122-4·8·3.5
Δ = 32
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:

$\sqrt{\Delta}=\sqrt{32}=\sqrt{16*2}=\sqrt{16}*\sqrt{2}=4\sqrt{2}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-4\sqrt{2}}{2*8}=\frac{12-4\sqrt{2}}{16}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+4\sqrt{2}}{2*8}=\frac{12+4\sqrt{2}}{16}
$