X^2+10x+1=40

Solution for X^2+10x+1=40 equation:

X^2+10X+1=40

We move all terms to the left:

X^2+10X+1-(40)=0

We add all the numbers together, and all the variables

X^2+10X-39=0

a = 1; b = 10; c = -39;
Δ = b2-4ac
Δ = 102-4·1·(-39)
Δ = 256
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$X_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{256}=16$
$X_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-16}{2*1}=\frac{-26}{2}
=-13
$
$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+16}{2*1}=\frac{6}{2}
=3
$