4x^2+42x-8=0

Solution for 4x^2+42x-8=0 equation:

4x^2+42x-8=0

a = 4; b = 42; c = -8;
Δ = b2-4ac
Δ = 422-4·4·(-8)
Δ = 1892
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:

$\sqrt{\Delta}=\sqrt{1892}=\sqrt{4*473}=\sqrt{4}*\sqrt{473}=2\sqrt{473}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(42)-2\sqrt{473}}{2*4}=\frac{-42-2\sqrt{473}}{8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(42)+2\sqrt{473}}{2*4}=\frac{-42+2\sqrt{473}}{8}
$