x^2+(x+16)=42^2

Solution for x^2+(x+16)=42^2 equation:

x^2+(x+16)=42^2

We move all terms to the left:

x^2+(x+16)-(42^2)=0

We add all the numbers together, and all the variables

x^2+(x+16)-1764=0

We get rid of parentheses

x^2+x+16-1764=0

We add all the numbers together, and all the variables

x^2+x-1748=0

a = 1; b = 1; c = -1748;
Δ = b2-4ac
Δ = 12-4·1·(-1748)
Δ = 6993
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:

$\sqrt{\Delta}=\sqrt{6993}=\sqrt{9*777}=\sqrt{9}*\sqrt{777}=3\sqrt{777}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-3\sqrt{777}}{2*1}=\frac{-1-3\sqrt{777}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+3\sqrt{777}}{2*1}=\frac{-1+3\sqrt{777}}{2}
$