7x^2+42x-161=0

Solution for 7x^2+42x-161=0 equation:

7x^2+42x-161=0

a = 7; b = 42; c = -161;
Δ = b2-4ac
Δ = 422-4·7·(-161)
Δ = 6272
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:

$\sqrt{\Delta}=\sqrt{6272}=\sqrt{3136*2}=\sqrt{3136}*\sqrt{2}=56\sqrt{2}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(42)-56\sqrt{2}}{2*7}=\frac{-42-56\sqrt{2}}{14}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(42)+56\sqrt{2}}{2*7}=\frac{-42+56\sqrt{2}}{14}
$