10x^2-28x-48=0

Solution for 10x^2-28x-48=0 equation:

10x^2-28x-48=0

a = 10; b = -28; c = -48;
Δ = b2-4ac
Δ = -282-4·10·(-48)
Δ = 2704
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{2704}=52$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-28)-52}{2*10}=\frac{-24}{20}
=-1+1/5
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-28)+52}{2*10}=\frac{80}{20}
=4
$