X^2+42x-20=0

Solution for X^2+42x-20=0 equation:

X^2+42X-20=0

a = 1; b = 42; c = -20;
Δ = b2-4ac
Δ = 422-4·1·(-20)
Δ = 1844
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:

$\sqrt{\Delta}=\sqrt{1844}=\sqrt{4*461}=\sqrt{4}*\sqrt{461}=2\sqrt{461}$
$X_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(42)-2\sqrt{461}}{2*1}=\frac{-42-2\sqrt{461}}{2}
$
$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(42)+2\sqrt{461}}{2*1}=\frac{-42+2\sqrt{461}}{2}
$