2x^2+20x-142=0

Solution for 2x^2+20x-142=0 equation:

2x^2+20x-142=0

a = 2; b = 20; c = -142;
Δ = b2-4ac
Δ = 202-4·2·(-142)
Δ = 1536
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:

$\sqrt{\Delta}=\sqrt{1536}=\sqrt{256*6}=\sqrt{256}*\sqrt{6}=16\sqrt{6}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-16\sqrt{6}}{2*2}=\frac{-20-16\sqrt{6}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+16\sqrt{6}}{2*2}=\frac{-20+16\sqrt{6}}{4}
$