7x^2-42x-8=0

Solution for 7x^2-42x-8=0 equation:

7x^2-42x-8=0

a = 7; b = -42; c = -8;
Δ = b2-4ac
Δ = -422-4·7·(-8)
Δ = 1988
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:

$\sqrt{\Delta}=\sqrt{1988}=\sqrt{4*497}=\sqrt{4}*\sqrt{497}=2\sqrt{497}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-42)-2\sqrt{497}}{2*7}=\frac{42-2\sqrt{497}}{14}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-42)+2\sqrt{497}}{2*7}=\frac{42+2\sqrt{497}}{14}
$