X^2+42x+48=0

Solution for X^2+42x+48=0 equation:

X^2+42X+48=0

a = 1; b = 42; c = +48;
Δ = b2-4ac
Δ = 422-4·1·48
Δ = 1572
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:

$\sqrt{\Delta}=\sqrt{1572}=\sqrt{4*393}=\sqrt{4}*\sqrt{393}=2\sqrt{393}$
$X_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(42)-2\sqrt{393}}{2*1}=\frac{-42-2\sqrt{393}}{2}
$
$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(42)+2\sqrt{393}}{2*1}=\frac{-42+2\sqrt{393}}{2}
$