X2+10x=42

Solution for X2+10x=42 equation:

X2+10X=42

We move all terms to the left:

X2+10X-(42)=0

We add all the numbers together, and all the variables

X^2+10X-42=0

a = 1; b = 10; c = -42;
Δ = b2-4ac
Δ = 102-4·1·(-42)
Δ = 268
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:

$\sqrt{\Delta}=\sqrt{268}=\sqrt{4*67}=\sqrt{4}*\sqrt{67}=2\sqrt{67}$
$X_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-2\sqrt{67}}{2*1}=\frac{-10-2\sqrt{67}}{2}
$
$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+2\sqrt{67}}{2*1}=\frac{-10+2\sqrt{67}}{2}
$