5x^2-42x-92=0

Solution for 5x^2-42x-92=0 equation:

5x^2-42x-92=0

a = 5; b = -42; c = -92;
Δ = b2-4ac
Δ = -422-4·5·(-92)
Δ = 3604
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:

$\sqrt{\Delta}=\sqrt{3604}=\sqrt{4*901}=\sqrt{4}*\sqrt{901}=2\sqrt{901}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-42)-2\sqrt{901}}{2*5}=\frac{42-2\sqrt{901}}{10}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-42)+2\sqrt{901}}{2*5}=\frac{42+2\sqrt{901}}{10}
$