2x^2-28x-48=0

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

2x^2-28x-48=0

a = 2; b = -28; c = -48;
Δ = b2-4ac
Δ = -282-4·2·(-48)
Δ = 1168
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:

$\sqrt{\Delta}=\sqrt{1168}=\sqrt{16*73}=\sqrt{16}*\sqrt{73}=4\sqrt{73}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-28)-4\sqrt{73}}{2*2}=\frac{28-4\sqrt{73}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-28)+4\sqrt{73}}{2*2}=\frac{28+4\sqrt{73}}{4}
$