x^2+18x+40=10

Solution for x^2+18x+40=10 equation:

X^2+18X+40=10

We move all terms to the left:

X^2+18X+40-(10)=0

We add all the numbers together, and all the variables

X^2+18X+30=0

a = 1; b = 18; c = +30;
Δ = b2-4ac
Δ = 182-4·1·30
Δ = 204
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:

$\sqrt{\Delta}=\sqrt{204}=\sqrt{4*51}=\sqrt{4}*\sqrt{51}=2\sqrt{51}$
$X_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(18)-2\sqrt{51}}{2*1}=\frac{-18-2\sqrt{51}}{2}
$
$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(18)+2\sqrt{51}}{2*1}=\frac{-18+2\sqrt{51}}{2}
$