8x^2-8x-42=0

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

8x^2-8x-42=0

a = 8; b = -8; c = -42;
Δ = b2-4ac
Δ = -82-4·8·(-42)
Δ = 1408
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:

$\sqrt{\Delta}=\sqrt{1408}=\sqrt{64*22}=\sqrt{64}*\sqrt{22}=8\sqrt{22}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-8)-8\sqrt{22}}{2*8}=\frac{8-8\sqrt{22}}{16}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-8)+8\sqrt{22}}{2*8}=\frac{8+8\sqrt{22}}{16}
$