0.75x^2-12x+40=0

Solution for 0.75x^2-12x+40=0 equation:

0.75x^2-12x+40=0

a = 0.75; b = -12; c = +40;
Δ = b2-4ac
Δ = -122-4·0.75·40
Δ = 24
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:

$\sqrt{\Delta}=\sqrt{24}=\sqrt{4*6}=\sqrt{4}*\sqrt{6}=2\sqrt{6}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-2\sqrt{6}}{2*0.75}=\frac{12-2\sqrt{6}}{1.5}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+2\sqrt{6}}{2*0.75}=\frac{12+2\sqrt{6}}{1.5}
$