5x^2-38x-48=0

Solution for 5x^2-38x-48=0 equation:

5x^2-38x-48=0

a = 5; b = -38; c = -48;
Δ = b2-4ac
Δ = -382-4·5·(-48)
Δ = 2404
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:

$\sqrt{\Delta}=\sqrt{2404}=\sqrt{4*601}=\sqrt{4}*\sqrt{601}=2\sqrt{601}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-38)-2\sqrt{601}}{2*5}=\frac{38-2\sqrt{601}}{10}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-38)+2\sqrt{601}}{2*5}=\frac{38+2\sqrt{601}}{10}
$