10x^2-48x+39=0

Solution for 10x^2-48x+39=0 equation:

10x^2-48x+39=0

a = 10; b = -48; c = +39;
Δ = b2-4ac
Δ = -482-4·10·39
Δ = 744
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:

$\sqrt{\Delta}=\sqrt{744}=\sqrt{4*186}=\sqrt{4}*\sqrt{186}=2\sqrt{186}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-48)-2\sqrt{186}}{2*10}=\frac{48-2\sqrt{186}}{20}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-48)+2\sqrt{186}}{2*10}=\frac{48+2\sqrt{186}}{20}
$